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p. No prime
1. We will show that there is a prime p > N such that p = 1 (mod 4). Let PROOF.
m = (N!)2 + 1.
Note that m is odd, m > 1. Let p be the smallest prime factor of m. None of the numbers 2,3, . . . , N divides m, so p > N. Also, we have (N !)’ =  1 (mod p). Raising both members to the (p  1)/2 power we find (N !)P ’ E ( l)(P ‘)I2 (mod p). But (N !)“ i = 1 (mod p) by the EulerFermat
theorem, so
( 1)(p1)/2 = 1 (mod p). Now the difference ( l)(p 1)/2  1 is either 0 or  2, and it cannot be  2, because it is divisible by p, so it must be 0. That is, (
l)(P
I)/2
=
1.
But this means that (p  1)/2 is even, so p = 1 (mod 4). In other words, we have shown that for each integer N > 1 there is a prime p > N such that p = 1 (mod 4). Therefore there are infinitely many primes of the form 4n + 1. 0 Simple arguments like those just given for primes of the form 4n  1 and 4n + 1 can also be adapted to treat other special arithmetic progressions, 147
7: Dirichlet’s
theorem
on primes in arithmetic
progressions
such as 5n  1,8n  1,8n  3 and 8n + 3 (see Sierpinski [67]), but no one has yet found such a simple argument that works for the general progression kn + h.
7.3 The plan of the proof of Dirichlet’s theorem In Theorem 4.10 we derived the asymptotic formula
p;x +
(2)
= log x + O(l),
where the sum is extended over all primes p I x. We shall prove Dirichlet’s theorem as a consequence of the following related asymptotic formula. Theorem 7.3 Ifk
> 0 and (h, k) = 1 we haue,for
(3) psh
IT p5.X
loi3P =
&
all x > 1,
log x + O(l),
(modk)
where the sum is extended over those primes p I x which are congruent to h mod k.
Since log x + co as x + co this relation implies that there are infinitely many primes p = h (mod k), hence infinitely many in the progression nk+h,n=0,1,2
,...
Note that the principal term on the right of (3) is independent of h. Therefore (3) not only implies Dirichlet’s theorem but it also shows that the primes in each of the q(k) reduced residue classesmod k make the same contribution to the principal term in (2). The proof of Theorem 7.3 will be presented through a sequence of lemmas which we have collected together in this section to reveal the plan of the proof. Throughout the chapter we adopt the following notation. The positive integer k represents a fixed modulus, and h is a fixed integer relatively prime to k. The q(k) Dirichlet characters mod k are denoted by xl,
x2
> . . . 3 &p(k)
with x1 denoting the principal character. For x # x1 we write L(1, x) and L’(1, x) for the sums of the following series: L(l,x)
= f x(n), n=l n
L’(l, x) = _ f n=l
148
xyg
n.
7.3: The plan of the proof of Dirichlet’s theorem
The convergence of each of these series was shown in Theorem 6.18. Moreover, in Theorem 6.20 we proved that L(1, x) # 0 if x is realvalued. The symbol p denotes a prime, and I,< x denotes a sum extended over all primes p I x. Lemma 7.4 For x > 1 we have
c 1% p=P $)
pzh
log x + $)
PSX (modk)
It is clear that Lemma 7.4 will imply Theorem 7.3 if we show that 1 x(P)log P = O(1) PSX P
(4)
for each x # x1. The next lemma expresses this sum in a form which is not extended over primes. Lemma 7.5 For x > 1 and 1( # xl we have
= L’(l, x)c @p + O(l). c x(p)log “2.x Pp a5.x Therefore Lemma 7.5 will imply (4) if we show that
(5)
&Mn)O(l). cp= “2X
This, in turn, will be deduced from the following lemma. Lemma 7.6 For x > 1 and x # x1 we have (6)
jJ,
x) 1
q
=
O(l).
n<x
If L(1, x) # 0 we can cancel L(1, x) in (6) to obtain (5). Therefore, the proof of Dirichlet’s theorem depends ultimately on the nonvanishing of L(1, x) for all x # x1. As already remarked, this was proved for real x # x1 in Theorem 6.20, so it remains to prove that L(1, 1) # 0 for all x # x1 which take complex as well as real values. For this purpose we let N(k) denote the number of nonprincipal characters x mod k such that L(1, x) = 0. If L(1, x) = 0 then L(l,X) = 0 and x # 1 since x is not real. Therefore the characters x for which L(1, x) = 0 occur in conjugate pairs, so N(k) is eoen. Our goal is to prove that N(k) = 0, and this will be deduced from the following asymptotic formula. 149
7: Dirichlet’s
theorem
on primes in arithmetic
progressions
Lemma 7.7 For x > 1 we have
?l ~51
= 1,(;‘k’ logx + O(1).
(modk)
If N(k) # 0 then N(k) 2 2 since N(k) is even, hence the coefficient of log x in (7) is negative and the right member + cc as x + co. This is a contradiction since all the terms on the left are positive. Therefore Lemma 7.7 implies that N(k) = 0. The proof of Lemma 7.7, in turn, will be based on the following asymptotic formula. Lemma 7.8 If x # x1 and L(1, x) = 0 we haue
L’(l,2)n<x 1 ep =log x + O(1). 7.4 Proof of Lemma 7.4 To prove Lemma 7.4 we begin with the asymptotic formula mentioned earlier,
& Ey = log x
(2)
+ O(1)
and extract those terms in the sum arising from primes p = h (mod k). The extraction is done with the aid of the orthogonality relation for Dirichlet characters, as expressed in Theorem 6.16: o?(k)
q(k)
r?lxr(m)Zr(n) = 0
if m = n (mod k), if m + n (mod k).
This is valid for (n, k) = 1. We take m = p and n = h, where (h, k) = 1, then multiply both members by p ’ log p and sum over all p I x to obtain
In the sum on the left we isolate those terms involving only the principal character x1 and rewrite (8) in the form (9)
q(k)
c p2.X pEh
+
= Xl(h) 1 P5.X
(modk)
Now Xl(h) = 1 and xi@) = 0 unless (p, k) = 1, in which case xi@) = 1. Hence the first term on the right of (9) is given by
(P,k)= 1
150
Plk
7.5: Proof of Lemma
7.5
since there are only a finite number of primes which divide k. Combining (10) with (9) we obtain @)
p1% P = 1 !t%! p5.x P
1
P2.X
p~h
+ ‘fj,(h)
c
r=2
PSX
“‘p
’ + O(1).
(modk)
Using (2) and dividing by q(k) we obtain Lemma 7.4.
cl
7.5 Proof of Lemma 7.5 We begin with the sum
x(n)W
Cy? nSx where A(n) is Mangoldt’s function, and express this sum in two ways. First we note that the definition of A(n) gives us
xW(4
cy= “SX
& *$I, xw;:g P”5.X We separate the terms with a = 1 and write
(11)
x(n)Nn) cy=
“2.x
1 x(Pbg P PCX
p.
P + c f x(p”Yog P psx a=2 Pa . P’2.X
The second sum on the right is majorized by
so (11) gives us c x(PYog P x(nPW) = Jx y+ O(l). P p5.X
(12)
Now we recall that A(n) = ‘&
P(d)log(n/d), hence
In the last sum we write n = cd and use the multiplicative to obtain
property of x
(13)
Since x/d 2 1, in the sum over c we may use formula (10) of Theorem 6.18 to obtain
xw% c cy=
L’(l,x)
+ 0
.
csxld
151
7: Dirichlet’s
theorem
on primes in arithmetic
progressions
Equation (13) now becomes (14)
The sum in the Oterm is ; d&(log x  log d) = 1 [x]log x  c log d = O(1) X
d<x
>
since 1 log d = log[x]!
= x log x + O(x).
d<x
Therefore (14) becomes L’(l,
x) 1
q
+ O(1)
d<x
which, with (12), proves Lemma 7.5.
Cl
7.6 Proof of Lemma 7.6 We use the generalized Mobius inversion formula proved in Theorem 2.23 which states that if c1is completely multiplicative we have (15)
G(x) = c cr(n)F x “5X
if, and only if, F(x) = c p(n)cc(n)G f . nsx
0
0
We take a(n) = x(n) and F(x) = x to obtain x =
(16)
C hUW n<x
t 0
where G(x) = 1 x(n) X = x 1 q. “5.x nbx By Equation (9) of Theorem 6.18 we can write G(x) = xL(1, x) + O(1). Using this in (16) we find x = 1 @)x(n)
“2X
i
X L(1, 2) + O(1) = xL(1, x) 1 F
I
Now we divide by x to obtain Lemma 7.6. 152
“lx
+ O(X). 0
7.8: Proof of Lemma
7.7
7.7 Proof of Lemma 7.8 We prove Lemma 7.8 and then use it to prove Lemma 7.7. Once again we make use of the generalized Mobius inversion formula (15). This time we take F(x) = x log x to obtain x log x = 1 &r)~(n)G ; nsx 0
(17)
where G(x) =
2 x(n); nsx
log ; = x log x 1 $) n<x
 x 1 ‘@)F “5X
n.
Now we use formulas (9) and (10) of Theorem 6.18 to get
= xL’(1, x) + O(log x) since we are assuming that L(1, x) = 0. Hence (17) gives us x log x = 1 &)x(n) ; L’(1, x) + 0 log ; “5X i ( I AM4 = xL’(1, x) 1 ~ + 0 1 (log x  log n) . nsx n ( n<x > We have already noted that the Oterm on the right is O(x) (see the proof of Lemma 7.5). Hence we have &)x(n) ‘(1, x) 1 ~ + O(x), nsx n and when we divide by x we obtain Lemma 7.8. x log x =
XL
0
7.8 Proof of Lemma 7.7 We use Lemma 7.4 with h = 1 to get
pnl
(modk)
In the sum over p on the right we use Lemma 7.5 which states that c PSX
LiP)log
P P
=
L’(l,
x,)
1
!q)
+
O(1).
“2X
153
7: Dirichlet’s
theorem
on primes in arithmetic
progressions
If L(1, x,) # 0, Lemma 7.6 shows that the right member of (18) is O(1). But if L(1, x,) = 0 then Lemma 7.8 implies L’(l,
x,) 1 !y
“SX Therefore the sum on the right of (18) is
= log x + O(1).
$) {W)lOg x + O(l)), so (18) becomes 1 pal
%
= l ,(Y
log x + O(1).
PSX (modk)
This proves Lemma 7.7 and therefore also Theorem 7.3.
cl
As remarked earlier, Theorem 7.3 implies Dirichlet’s theorem: Theorem 7.9 If k > 0 and (h, k) = 1 there are injinitely arithmetic progression nk + h, n = 0, 1, 2, . . .
many primes in the
7.9 Distribution of primes in arithmetic progressions Ifk > Oand(a,k)
= 1,let %b) =
p;x PI
II (mod
1. k)
The function n,(x) counts the number of primes IX in the progression nk+a,n=0,1,2 ,... Dirichlet’s theorem shows that z,(x) + co as x + co. There is also a prime number theorem for arithmetic progressions which states that (19)
44
n&x)L
q(k)
X
q(k) log x
as ’ + O”’
if (a, k) = 1. A proof of (19) is outlined in [44]. The prime number theorem for progressions is suggested by the formula of Theorem 7.3, 1% P
ET
PSX prh
= & log x + O(1).
(modk)
Since the principal term is independent of h, the primes seem to be equally distributed among the q(k) reduced residue classes mod k, and (19) is a precise statement of this fact. 154
Exercises for Chapter
7
We conclude this chapter by giving an alternate formulation of the prime number theorem for arithmetic progressions. Theorem 7.10 Ifthe
relation
Q(X)  w as x + co dk)
(20)
holds for every integer a relatively prime to k, then %W N %(4 asx+oo whenever (a, k) = (b, k) = 1. Conversely, (21) implies (20).
(21)
PROOF.It is clear that (20) implies (21). To prove the converse we assume (21) and let A(k) denote the number of primes that divide k. If x > k we have n(x) = 1 1 = A(k) + c 1 P5.X P2.X PXk = A(k) +
i a=1 (a,k)=l
Therefore
1 pza
1 = A(k) +
5
q(x).
a=1 (a, k) = 1
p5.X (modk)
44 A(k) %(X) (a,k)=l
By (21) each term in the sum tends to 1 as x + cc so the sum tends to q(k). Hence 44 44 q(k) __asxrco. %(X) %(4 But A(k)/q,(x) f 0 so 71(x)/q,(x) + q(k), which proves (20). 0
Exercises for Chapter 7 In Exercises 1 through 4, h and k are given positive integers, (h, k) = 1, and A(h, k) is the arithmetic progression A(h, k) = {h + kx:x = 0, 1, 2, . . .}. Exercises 1 through 4 are to be solved without using Dirichlet’s theorem. 1. Prove that, for every integer n 2 1, A@, k) contains infinitely prime to n. 2. Prove that A@, k) contains i#j.
an infinite
many numbers
relatively
subset {a,, a2,. . .} such that (ai, aj) = 1 if
3. Prove that A&, k) contains an infinite subset which forms a geometric progression (a set of numbers of the form ar”, n = 0, 1, 2, .). This implies that A@, k) contains infinitely many numbers having the same prime factors.
155
7: Dirichlet’s
theorem
on primes in arithmetic
progressions
4. Let S be any infinite subset of A(h, k). Prove that for every positive integer n there is a number in A(h, k) which can be expressed as a product of more than n different elements of S. 5. Dirichlet’s theorem implies the following statement: If h and k > 0 are any two integers with (h, k) = 1, then there exists at least one prime number of the form kn + h. Prove that this statement also implies Dirichlet’s theorem. 6. If (h, k) = 1, k > 0, prove that there is a constant that, if x 2 2,
A (depending
7. Construct an infinite set S of primes with the following property: then ($p  I), $(q  1)) = (p, q  l)‘= (p  1, q) = 1.
on h and on k) such
If p E S and q E S
8. Letfbe an integercoefficient polynomial ofdegree n 2 1 with the following property: For each prime p there exists a prime q and an integer m such that f(p) = q”. Prove that q = p, m = n and f(x) = x” for all x. [Hint: If q # p then q”‘+’ divides f(p+tq”+‘)f(p)foreacht= 1,2,...]
156
Periodic Arithmetical Functions and Gauss Sums
8
8.1 Functions periodic modulo k Let k be a positive integer. An arithmetical functionf with period k (or periodic modulo k) if
is said to be periodic
f(n I 4 = f(n) for all integers n. If k is a period so is mk for any integer m > 0. The smallest positive period off is called the fundamental period. Periodic functions have already been encountered in the earlier chapters. For example, the Dirichlet characters mod k are periodic mod k. A simpler example is the greatest common divisor (n, k) regarded as a function of n. Periodicity enters through the relation (n + k, k) = (n, k).
Another example is the exponential function f(n) = e2nimW where m and k are fixed integers. The number eZnimikis a kth root of unity andf(n) is its nth power. Any finite linear combination of such functions, say
is also periodic mod k for every choice of coefficients c(m). Our first goal is to show that every arithmetical function which is periodic mod k can be expressed as a linear combination of this type. These sums are called$nite Fourier series. We begin the discussion with a simple but important example known as the geometric sum. 157
8 : Periodic arithmetical functions and Gauss sums
Theorem 8.1 For jxed
k 2 1 let g(n)
kl c
=
,+imnlk.
m=O
0 k
s(n) =
PROOF.
ifk,+‘n, if kin.
Since g(n) is the sum of terms in a geometric
progression,
kl s(n)
=
1 m=O
xm,
where x = ezninik,we have
I
Xk  1 ifx # 1, g(n) = . x  1 k ifx = 1. But xk = 1, and x = 1 if and only if kin, so the theorem is proved.
0
8.2 Existence of finite Fourier series for periodic arithmetical functions We shall use Lagrange’s polynomial interpolation formula to show that every periodic arithmetical function has a finite Fourier expansion. Theorem 8.2 Lagrange’s interpolation theorem. Let zo, zl, . . , zk 1 be k distinct complex numbers, and let wo, wl, . . . , wk 1 be k complex numbers which need not be distinct. Then there is a unique polynomial P(z) of degree I k  1 such that fY%l) = wnl for m = 0, 1,2, . . . , k  1.
The required polynomial P(z), called the Lagrange interpolation polynomial, can be constructed explicitly as follows. Let
PROOF.
A(z) = (z  zo)(z  zl) . . . (z  Zk 1) and let
A,(z) = ~44 zzz,’ Then A,(z) is a polynomial of degree k  1 with the following properties: &(z,) 158
Z 0,
A,(Zj) = 0
ifj # m.
8.2: Existence
of finite Fourier seriesfor periodic arithmetical functions
Hence A,(z)/A,(z,) is a polynomial of degree k  1 which vanishes at each Zj for j # m, and has the value 1 at z,. Therefore the linear combination
is a polynomial of degree I k  1 with P(zj) = wj for each j. If there were another such polynomial, say Q(z), the difference P(z)  Q(z) would vanish at k distinct points, hence P(z) = Q(z) since both polynomials have degree 1.
1 p(d) =: [A]={0 dlh k)
Hence d,pkdp(!j) “3
== ,zdk
e2nimnlk = ck(n).
(m,k)=l
8.4 Multiplicative sums S&Z)
properties of the
Theorem 8.7 Let Sk(n)
=
1
f(&
dlh k)
where f and g are multiplicative. (9) 162
s,,&b)
= s,(&,(b)
Then we have whenever (a, k) = (b, m) = 1.
Cl
8.4: Multiplicative properties of the sumss&z)
In particular,
we have s,(ab) = s,(a)
(10)
if(b, m) = 1,
and if (a, k) = 1.
s&a) = s,(a)g(k)
(11)
PROOF.The relations (a, k) = (b, m) = 1 imply (see Exercise 1.24) W, 4
= (a, m)(k b)
with (a, m) and (b, k) relatively prime. Therefore
Writing d = d,d, in the last sum we obtain
%&b)= dll(o,m) 1 d#‘,k) c f(dAg This proves (9). Taking k = 1 in (9) we get s,(ab) = s,(a)sI(b)
since s,(b) = f(l)g(l)
= s,,,(a)
= 1. This proves (10). Taking b = 1 in (9) we find %kta)
=
sm(a)sk(l)
=
G&W
since sk(1) = f( l)g(k) = g(k). This proves (11).
0
EXAMPLE For Ramanujan’s sum we obtain the following
multiplicative
properties : whenever (a, k) = (b, m) = 1, whenever (b, m) = 1,
Cm,@‘) = d&,(b) c&b) = c,(a) and &k(a)
=
whenever (a, k) = 1.
c,&Mk)
Sometimes the sums Sk(n) can be evaluated in terms of the Dirichlet convolutionf * g. In this connection we have: Theorem 8.8 Let f be completely multiplicative, h is multiplicative. and let
Assume that
Sk(n)
=
f
and let g(k) = p(k)h(k), where (p) # 0 and f (p) # h(p) for all primes p,
1 f (dig dlh k)
. 163
8 : Periodic
arithmetical
functions
and Gauss sums
Then we have s
(n)
=
Wg(N)
k J’(N)
’
where F = f * g and N = k/(n, k).
PROOF.First we note that
F(k) = c f(d)p k h k  c f k 44W = f(k); dlk
cd)
cd)
= /(k);(l

 $j
>
d,k
a(d)#
cd)
.
Next, we write a = (n, k), so that k = aN. Then we have Sk@)
=
1 dlo
f@)~
2 (“>
h
j (“1
=
f (& 7
c dla
(““>
h d(““1
Now p(Nd) = p(N)p(d) if (N, d)  1, and p(Nd) = 0 if (N, d) > 1, so the last equation gives us sk(n)
=
&WN)
f
c
db
z
N)h(d)
f(a)p(N)h(N)
=
0
(N,d)=
‘i p(d) E 40 (N,d)=l
1
= f(4NW4N)~ PXN
F(k) f(N) WMWW) =~GMWW)~OF(~~ = F(N)
FW&V =F(Ni.
EXAMPLE For Ramanujan’s sum we obtain the following simplification:
ck(n)
164
=
cp(k)p(Nh(N)
=
0
8.5 : Gauss sums associated
with Dirichlet
characters
8.5 Gauss sums associated with Dirichlet characters Definition For any Dirichlet character x mod k the sum G(n, x) = i &t)e2nimn’k m=l
is called the Gauss sum associated with x. If x = x1, the principal character mod k, we have x1(m) = 1 if (m, k) = 1, and x1(m) = 0 otherwise. In this case the Gauss sum reduces to Ramanujan’s sum : G(n, xl) =
i
e2nimn/k=
ck(n).
m=l (m,k)=l
Thus, the Gauss sums G(n, x) can be regarded as generalizations of Ramanujan’s sum. We turn now to a detailed study of their properties. The first result is a factorization property which plays an important role in the subsequent development. Theorem 8.9 If x is any Dirichlet
character mod k then
G(n, x) = z(n)G(l, x)
whenever (n, k) = 1.
PROOF. When (n, k) = 1 the numbers nr run through a complete residue system mod k with r. Also, 1x(n) 1’ = x(n)f(n) = 1 so
x(r) = i3nMnM) = dnkh). Therefore the sum defining G(n, x) can be written as follows: G(n, x) =
c
X(r)e2ninr’k = j(n)
rmodk
c
X(nr)e2ninr’k
rmodk
= r?(n)1 xbk 2nim’k= j(n)G(l, x). mmodk
q
This proves the theorem. Definition The Gauss sum G(n, x) is said to be separable if (12)
Gh xl = XWW, xl.
Theorem 8.9 tells us that G(n, x) is separable whenever n is relatively prime to the modulus k. For those integers n not relatively prime to k we have the following theorem. 165
8 : Periodic
arithmetical
functions
and Gauss sums
Theorem 8.10 If x is a character mod k the Gauss sum G(n., x) is separable for every n if and only if, G(n, x) == 0
whenever (n, k) > 1.
PROOF. Separability always holds if (n, k) = 1. But if (n, k) > 1 we have j(n) = 0 so Equation (12) holds if and only if G(n, x) = 0. cl
The next theorem gives an important consequence of separability. Theorem 8.11 Zf G(n, x) is separablefor (13) PROOF.
every n then
IGU, x)l2 = k.
We have 
IG(1, x)12 = G(l, x)G(l, x) = G(l, x) 1 j(m)e2”im~k m=l
= mclG(m, X)e2nim/k = mil .il X(r)e2nin1rlke2nimlk
= ilx(4
i: e2nim(rl)/k = Q(l)
= k,
In=1
since the last sum over m is a geometric sum which vanishes unless r
=
1.
8.6 Dirichlet characters with nonvanishing Gauss sums For every character x mod k we have seenthat G(n, x) is separable if (n, k) = 1, and that separability of G(n, x) is equivalent to the vanishing of G(n, x) for (n, k) > 1. Now we describe further properties of those characters such that G(n, x) = 0 whenever (n, k) > 1. Actually, it is simpler to lstudy the complementary set. The next theorem gives a necessary condition for G(n, 1) to be nonzero for (n, k) > 1. Theorem 8.12 Let x be a Dirichlet character mod k and assume that G(n, x) # 0 for some n satisfying (n, k) > 1. Then there exists a dicisor d of k, d < k, such that
(14)
x(a) = 1 wheneuer (a, k) = 1 and a z 1 (mod d).
For the given n, let 4 = (n, k) and let d = k/q. Then d I k and, since q > 1, we have d < k. Choose any a satisfying (a, k) = 1 and a = 1 (mod d). We will prove that x(a) = 1. PROOF.
166
8.7: Induced moduli and primitive characters
Since (a, k) = 1, in the sum defining G(n, x) we can replace the index of summation m by am and we find G(n, x) =
c
X(m)e2ninm’k =
mmodk
= x(a) 1
X(m)e2ninam’k.
mmodk
Since a = 1 (mod d) and d = k/q we can write a = 1 + (bk/q) for some integer b, and we have anm =+k
nm k
bknm = t qk
+ brim E t 4
(mod 1)
since q 1n. Hence e2ninrrm’k = e2ninm’kand the sum for G(n, x) becomes
Gh xl = x~~)m~dkx.(mk2ri”m/*= x(Wh
xl.
Since G(n, x) # 0 this implies x(a) = 1, as asserted.
Cl
The foregoing theorem leads us to consider those characters x mod k for which there is a divisor d < k satisfying (14). These are treated next.
8.7 Induced
moduli
and primitive
characters
of induced modulus Let x be a Dirichlet character mod k and let d be any positive divisor of k. The number d is called an induced modulus for x if we have
Definition
(15)
x(a) = 1 whenever (a, k) = 1 and a = 1 (mod d).
In other words, d is an induced modulus if the character x mod k acts like a character mod d on the representatives of the residue class f mod d which are relatively prime to k. Note that k itself is always an induced modulus for x. Theorem 8.13 Let x be a Dirichlet character modulus for x $ and only if, x = x1.
mod k. Then 1 is an induced
If x = x1 then x(a) = 1 for all a relatively prime to k. But since every a satisfies a = 1 (mod 1) the number 1 is an induced modulus. Conversely, if 1 is an induced modulus, then x(a) = 1 whenever (a, k) = 1, so x = x1 since x vanishes on the numbers not prime to k. 0
PROOF.
167
8 : Periodic arithmetical functions and Gausssums For any Dirichlet character mod k the modulus k itself is an induced modulus. If there are no others we call the character ,~imitiue. That is, we have: of primitive characters A Dirichlet character x mod k is said to be primitive mod k if it has no induced modulus d < k. In other words, x is primitive mod k if, and only if, for every divisor d of k, 0 < d < k, there exists an integer a = 1 (mod d), (a, k) = 1, such that x(a) # 1.
Definition
If k > 1 the principal character x1 is not primitive since it has 1 as an induced modulus. Next we show that if the modulus is prime every nonprincipal character is primitive. Theorem 8.14 Every nonprincipal character mod p.
character
x modulo a prime p is a primitive
PROOF. The only divisors of p are 1 and p so these are the only candidates for induced moduli. But if x # x1 the divisor 1 is not an induced modulus so x has no induced modulus 1. (b) G(n, x) is separable for every n. (4 IGU, x)l’ = k.
If G(n, x) # 0 for some n with (n, k) > 1 then Theorem 8.12 shows that x has an induced modulus d < k, so x cannot be primitive. This proves (a). Part (b) follows from (a) and Theorem 8.10. Part (c) follows from part (b) and Theorem 8.11. q PROOF.
Note. Theorem 8.15(b) shows that the Gauss sum G(n, x) is separable if x is primitive. In a later section we prove the converse. That is, if G(n, 1) is separable for every n then 2 is primitive. (SeeTheorem 8.19.)
8.8 Further
properties
of induced moduli
The next theorem refers to the action of x on numbers which are congruent modulo an induced modulus. Theorem 8.16 Let x be a Dirichlet character mod k and assume d 1k, d > 0. Then d is an induced modulus for x if, and only if,
(16) 168
x(a) = X(b)
whenever (a, k) = (b, k) = 1 and a z b (mod d).
8.8 : Further
properties
of induced
moduli
PROOF. If (16) holds then d is an induced modulus since we may choose b = 1
and refer to Equation (15). Now we prove the converse. Choose a and b so that (a, k) = (b, k) = 1 and a = b (mod d). We will show that x(a) = X(b). Let a’ be the reciprocal of u mod k, au’  1 (mod k). The reciprocal exists because (a, k) = 1. Now au’  1 (mod d) since d I k. Hence ~(uu’) = 1 since d is an induced modulus. But au’ = bu’ 3 1 (mod d) because a = b (mod d), hence x(au’) = X(bu’), so
But ~(a’) # 0 since x(u)x(u’) = 1. Canceling ~(a’) we find x(u) = X(b), and this completes the proof. 0 Equation (16) tells us that x is periodic mod d on those integers relatively prime to k. Thus x acts very much like a character mod d. To further explore this relation it is worthwhile to consider a few examples. EXAMPLE 1 The following table describes one of the characters x mod 9. n
1
x(n)
1
2 1
3
4
5
0
1
1
6
7
0
1
8 1
9 0
We note that this table is periodic modulo 3 so 3 is an induced modulus for x. In fact, x acts like the following character + modulo 3 : n
1
2
3
$(n)
1
 1
0
Since x(n) = I&) for all n we call x an extension of t+Q. It is clear that whenever x is an extension of a character $ modulo d then d will be an induced modulus for x. EXAMPLE 2 Now we examine one of the characters x modulo 6: n
1
2
3
4
5
6
x(n)
1
0
0
0
 1
0
In this case the number 3 is an induced modulus because x(n) = 1 for all n = 1 (mod 3) with (n, 6) = 1. (There is only one such n, namely, n = 1.) 169
8 : Periodic arithmetical functions and Gauss sums
However, x is not an extension of any character II/ modulo 3, because the only characters modulo 3 are the principal character +i, ,given by the table: n
1
2
3
and the character $ shown in Example 1. Since x(2) = 0 it cannot be an extension of either + or $,. These examples shed some light on the next theorem. Theorem 8.17 Let x be a Dirichlet Then the following
character modulo k and assume d 1k, d > 0. two statements are equivalent:
(a) d is an induced modulus for x. (b) There is a character II/ modulo d such that (17) where x1 is the principal
x(n) = rl/(nh(n)
for all 4
character modulo k.
PROOF. Assume (b) holds. Choose n satisfying (n, k) = 1, n = 1 (mod d). Then x1(n) = IC/(n) = I so x(n) = 1 and hence d is an induced modulus. Thus, (b) implies (a). Now assume (a) holds. We will exhibit a character I++modulo d for which (17) holds. We define $(n) as follows: If (n, d) > 1, let I&) = 0. In this case we also have (n, k) > 1 so (17) holds because both members are zero. Now suppose (n, d) = 1. Then there exists an integer m such that m E n (mod d), (m, k) = 1. This can be proved immediately with Dirichlet’s theorem. The arithmetic progression xd + n contains infinitely many primes. We choose one that does not divide k and call this m. However, the result is not that deep; the existence of such an m can easily be e;stablished without using Dirichlet’s theorem. (See Exercise 8.4 for an alternate proof.) Having chosen m, which is unique modulo d, we define
ti(n) = x(m). The number e(n) is welldefined because x takes equal values at numbers which are congruent modulo d and relatively prime to k. The reader can easily verify that x is, indeed, a character mod d. We shall verify that Equation (17) holds for all n. If (n, k) = 1 then (n, d) = 1 so $(n) = x(m) for some m E n (mod d). Hence, by Theorem 8.16, x(n) = x(m) = IL(n) = Wkl(n) since x1(n) = 1. If (n, k) > 1, then x(n) = xl(n) = 0 and both members of (17) are 0. Thus, 0 (17) holds for all n. 170
8.10: Primitive charactersand separableGausssums
8.9 The conductor
of a character
Let x be a Dirichlet character mod k. The smallest induced modulus d for x is called the conductor of x.
Definition
Theorem8.18
Every Dirichlet
(18)
character x mod k can be expressed as a product,
x(n) = Il/(nkl(n)
where x1 is the principal character modulo the conductor of 1+9.
for all n,
mod k and $ is a primitive
character
PROOF.Let d be the conductor of x. From Theorem 8.17 we know that x can be expressed as a product of the form (18), where II/ is a character mod d. Now we shall prove that $ is primitive mod d. We assume that $ is not primitive mod d and arrive at a contradiction. If I,+is not primitive mod d there is a divisor q of d, q < d, which is an induced modulus for $. We shall prove that this q, which divides k, is also an induced modulus for x, contradicting the fact that d is the smallest induced modulus for x. Choose n = 1 (mod q), (n, k) = 1. Then x(4 = vWxl(n) = 964 = 1 because q is an induced modulus for II/. Hence q is also an induced modulus for x and this is a contradiction. 0
8.10 Primitive characters Gauss sums
and separable
As an application of the foregoing theorems we give the following alternate description of primitive characters. Theorem 8.19 Let x be a character only if, the Gauss sum G(n, x) =
mod k. Then x is primitive
1
mod k if, and
X(m)e2zimn’k
mmodk
is separable for every n.
PROOF.If x is primitive, then G(n, x) is separable by Theorem 8.15(b). Now we prove the converse. Because of Theorems 8.9 and 8.10 it suffices to prove that if x is not primitive mod k then for some r satisfying (r, k) > 1 we have G(r, x) # 0. Suppose, then, that x is not primitive mod k. This implies k > 1. Then x has a conductor d < k. Let r = k/d. Then (r, k) > 1 and we shall prove that 171
8 : Periodic
arithmetical
functions
and Gauss sums
G(r, x) # 0 for this r. By Theorem 8.18 there exists a primitive character $ mod d such that x(n) = r/+)x1(n) for all n. Hence we can write G(r, x) =
1
t)(m)Xl(m)e2xirm’k= m,C, kWWnirmik
mmodk (m,k)=
=
mzdk
be+
(m,k)=
2nimld
=
g
1
1
mzdd
(m,d)=
$,(m)e2sWd,
1
where in the last step we used Theorem 5.33(a). Therefore we have G(r, x) = g
G( 1, II/).
But 1G(l, $)I2 = d by Theorem 8.15 (since $ is primitive mod d) and hence G(r, x) # 0. This completes the proof. q
8.11 The finite Fourier characters
series of the Dirichlet
Since each Dirichlet character x mod k is periodic mod k it has a finite Fourier expansion k
(19)
x(m) = 1 ak(n)e2nimn’k,
and Theorem 8.4 tells us that its coefficients are given by the formula uk(n) = t i X(m)e2”im”ik. ml The sum on the right is a Gauss sum G(  n, x) so we have (20)
a,‘(n) = ; G( n, X).
When x is primitive the Fourier expansion (19) can be expressed as follows : Theorem 8.20 The finite Fourier expansion of a primitive x mod k has the form (21)
x(m) = !$$
i ji(n)e2”imdk n 1
where (2a
Zk(X) = TG(17x) = 5
The numbers tk(x) have absolute value 1.
172
~~I~(m)e2”im~k.
Dirichlet
character
8.12: Pdya’s
inequality
for the partial
sums of primitive
characters
PROOF. Since x is primitive we have G( n, x) = j( n)G(l, x) and (20) implies L&I) = j(  n)G(l, x)/k. Therefore (19) can be written as
which is the same as (21). Theorem 8.11 shows that the numbers zk(x) have absolute value 1. 0
8.12 Pblya’s inequality for the partial sums of primitive characters The proof of Dirichlet’s theorem given in Chapter 7 made use of the relation I
m;~(m)
I
I dk)
which holds for any Dirichlet character x mod k and every real x 2 1. This cannot be improved when x = x1 because cm=, x1(m) = q(k). However, Polya showed that the inequality can be considerably improved when x is a primitive character. Theorem 8.21 Polya’s inequality. If x is any primitive for all x 2 1 we have
character
mod k then
(23) PROOF.
We express x(m) by its finite Fourier expansion, as given in Theorem
8.20
x(m) = ?$$! $ x(n)e2nimnlk, ”
1
and sum over all m I x to get
since X(k) = 0. Taking absolute values and multiplying by & we find
say, where
173
8 : Periodic
arithmetical
functions
and Gauss sums
Now f(k
_
n)
=
2
e
2nim(kN/k
=
1
m<x
e2nimn/k
=
f(n)
msx
so
(f(k  n) 1= 1f(n) I. Hence (24) can be written as (25)
Nowf(n)
is a geometric sum of the form f(n) = i
Y”
m=l
where r = [x] and y = e 2nin’k. Here y # 1 since 1 I n < k  1. Writing z = exinlk, we have y = z2 and ,z2 # 1 since n I k/2. Hence we have
NOW we use the inequality m/k to get
sin t 2 2t/x, valid for Cl 5 t < 7~12, with t =
Irk
Hence (25) becomes I k c &k/2
and this proves (23).
1c k log k, n 0
Note. In a later chapter we will prove that Polya’s inequality can be extended to any nonprincipal character. For nonprimitive characters it takes the form
(See Theorem 13.15.) 174
Exercises for Chapter
8
Exercises for Chapter 8 1. Let x = ezniin and prove that
2. Let ((x)) = x  [x]  $ if x is not an integer, and let ((x)) = 0 otherwise. Note that ((x)) is a periodic function of x with period 1. If k and n are integers, with n > 0, prove that k (0) n
=
3. Let ck(m) denote Ramanujan’s Mobius function.
in $rrcot
y
sin T
.
sum and let M(x) = cnSX p(n), the partial
sums of the
(a) Prove that k$lckh,
In particular,
=
1 dM(;). dim
=
dF
when n = m, we have k~,ckb,
dM(;). n
(b) Use (a) to deduce that M(m)
= m 1 y dim
i
c,(d).
k=l
(c) Prove that m$lck(m)
=
c
&(;)[j.
dlk
4. Let n, a, d be given integers with (a, d) = 1. Let m = a + qd where q is the product (possibly empty) of all primes which divide n but not a. Prove that m E a (mod d) 5. Prove that there exists no real primitive
and(m,n)
character x mod k if k = 2m, where m is odd.
6. Let 1 be a character mod k. If kl and k2 are induced (k,, k2), their gtd. 7. Prove that the conductor
= 1.
of 1 divides every induced
moduli modulus
for x prove that so too is for x.
In Exercises 8 through 12, assume that k = k,kz ... k,, where the positive integers ki are relatively prime in pairs: (ki, kj) = 1 if i # j. 8. (a) Given any integer a, prove that there is an integer ai such that ai = a (mod ki)
and ai = 1 (mod kj)
for all j # i.
175
8 : Periodic
arithmetical
functions
(b) Let x be a character
and Gauss sums
mod k. Define xi by the equation =
Idal
x(4X
where ai is the integer of part (a). Prove that xi is a character
mod ki.
9. Prove that every character x mod k can be factored uniquely form x = x1 x2 . . x,, where xi is a character mod ki. 10. Let f(x) denote the conductor that f(x) = f(xJ
of x. If x has the factorization
as a product
of the
in Exercise 9, prove
. fW
11. If 2 has the factorization
in Exercise 9, prove that for every integer a we have
G(aa X)= iolXi(3k G(aia Xi), where ai is the integer of Exercise 8. 12. If x has the factorization in Exercise 9, prove that x is primitive each xi is primitive mod ki. [Hint: Theorem 8.19.1 13. Let x be a primitive
character
mod k if, and only if,
mod k. Prove that if N < M we have
I.;+,$1
< A1
Jit
log k.
14. This exercise outlines a slight improvement in Polya’s inequality. of Theorem 8.21. After inequality (26) write
Show that the integral
is less than (k/n)log(sin(~/2k))
Refer to the proof
and deduce that
In&x:Ix(n) I 0, the sum G(k ; n) = i e2niWn r=1 is called a quadratic Gauss sum. Derive the following properties of quadratic Gauss sums : (a) G(k; mn) = G(km; n)G(kn; m) whenever (m, n) = 1. This reduces the study of Gauss sums to the special case G(k; p”), where p is prime. (b) Let p be an odd prime, p ,j’ k, tl 2 2. Prove that G(k; p”) = pG(k; pa‘) and deduce that
ai2 G(k;p”)
=
’ 1 p@ ‘)“G(k;
if t( is even, p)
if tl is odd.
Further properties of the Gauss sum G(k; p) are developed in the next chapter where it is shown that G(k; p) is the same as the Gauss sum G(k, x) associated with a certain Dirichlet character x mod p. (See Exercise 9.9.)
177
9
Quadratic Residues and the Quadratic Reciprocity Law
9.1 Quadratic
residues
As shown in Chapter 5, the problem of solving a polynomial congruence f(x) E 0 (mod m) can be reduced to polynomial congruences with prime moduli plus a set of linear congruences. This chapter is concerned with quadratic congruences of the form x2 z n (mod p)
(1)
where p is an odd prime and n f 0 (mod p). Since the modulus is prime we know that (1) has at most two solutions. Moreover, if x is a solution so is x, hence the number of solutions is either 0 or 2. If congruence (1) has a solution we say that n is a quadratic residue mod p and we write nRp. If (1) has no solution we say that n is a quadratic nonresidue mod p and we write r&p.
Definition
Two basic problems dominate the theory of quadratic residues: 1. Given a prime p, determine which n are quadratic residues mod p and which are quadratic nonresidues mod p. 2. Given n, determine those primes p for which n is a quadratic residue mod p and those for which n is a quadratic nonresidue mod p. We begin with some methods for solving problem 1. 178
9.2 : Legendre’s symbol and its properties
EXAMPLE To find the quadratic residues modulo 11 we square the numbers
1,2,...,
10 and reduce mod 11. We obtain
12 z 1,
22 s 4,
32 = 9,
42 = 5,
52 G 3 (mod 11).
It suffices to square only the first half of the numbers since 62 ZE(5)2 z 3,
72 z (4)2 E 5,...,
lo2 E ( 1)2 E 1 (mod 11).
Consequently, the quadratic residues mod 11 are 1, 3, 4, 5, 9, and the nonresidues are 2,6, 7, 8, 10. This example illustrates the following theorem. Theorem 9.1 Let p be an odd prime. Then every reduced residue system mod p contains exactly (p  1)/2 quadratic residues and exactly (p  1)/2 quadratic nonresidues mod p. The quadratic residues belong to the residue classes containing the numbers
pl 2 12, 22, 32,. . .) ~ ( 2 > . First we note that the numbers in (2) are distinct mod p. In fact, if x2 = y2 (mod p) with 1 < x I (p  1)/2 and 1 I y 5 (p  1)/2, then
PROOF.
(x  y)(x + y) 3 0 (mod p). But 1 < x + y < p so x  y c 0 (mod p), hence x = y. Since 0,  k)’ = k2 (mod p),
every quadratic residue is congruent mod p to exactly one of the numbers in (2). This completes the proof. cl The following brief table of quadratic residues R and nonresidues i? was obtained with the help of Theorem 9.1. p=3
p=5
p=l
p= 11
p= 13
R:
1
L4
1,2,4
1, 334, 59
1, 3,4,9, 10, 12
R:
2
2, 3
3, 5, 6
2, 6, 7, 8, 10
2, 5, 6, 7, 8, 11
9.2 Legendre’s symbol and its properties Let p be an odd prime. If n + 0 (mod p) we define Legendre’s symbol (n 1p) as follows : + 1 if nRp, (nip) = i  1 if r&p.
Definition
If n = 0 (mod p) we define (n 1p) = 0. 179
9: Quadratic residuesand the quadratic reciprocity law
EXAMPLES(~~P)= l,(m’Ip)
= 1,(7)11) = 1,(22[11)
= 0.
Note. Some authors write $ instead of (n Ip). 0 It is clear that (m Ip) = (n Ip) whenever m = n (mod p), so (n Ip) is a periodic function of n with period p. The little Fermat theorem tells us that ripl GE1 (mod p) if p ,j’ n. Since n p1
_
1 =
(n(Pw2
_
l)(n’P“12
+
1)
it follows that nCp 1)‘2 = + 1 (mod p). The next theorem tells us that we get + 1 if nRp and  1 if ni?p.Theorem9.2 Euler’s criterion. Let p be an odd prime. Then for all n we have (n Ip) = nCp 1)‘2 (mod p). PROOF. If n = 0 (mod p) the result is trivial since both members are congruent to 0 mod p. Now suppose that (nip) = 1. Then there is an x such that x2 = n (mod p) and hence a(P
1)/z
E
(~2)(P1)/2
=
xP1
E
1
=
(n
Ip)
(mod
p).
This proves the theorem if (n I p) = 1. Now suppose that (n Ip) =  1 and consider the polynomial f(x) = ,$1)/2  1. Sincef(x) has degree (p  1)/2 the congruence f(x) = 0 (mod p) has at most (p  1)/2 solutions. But the (p  1)/2 quadratic residues mod p are solutions so the nonresidues are not. Hence nCp‘)”
f 1 (mod p)
if(nlp) = 1.
But n(P l)j2 = + 1 (mod p) so nCp ‘)I2 =  1 = (n Ip) (mod p). This com0 pletes the proof.Theorem 9.3 Legendre’s symbol (n I p) is a completely multiplicative
function
of n.
PROOF.If plm or pin then plmn so (mnlp) = 0 and either (mlp) = 0 or (nip) = 0. Therefore (mnlp) = (mjp)(nlp) ifplm or pin. Ifp$mandp$nthenp,j’mnandwehave (mn Ip) = (mn)(“ ‘)I2 = rn(P ‘)i2n(P ‘)I2 = (m Ip)(n Ip) (mod p).
180
9.3: Evaluation
of ( 1 Jp) and (21~)
But each of (mn 1p), (m 1p) and (n 1p) is 1 or  1 so the difference
(mnld  (mIP)(nIP) is either 0,2, or  2. Since this difference is divisible by p it must be 0. Note. Since (n jp) is a completely multiplicative function of n which is periodic with period p and vanishes when p (n, it follows that (n Ip) = x(n), where x is one of the Dirichlet characters modulo p. The Legendre symbol is called the quadratic character mod p.
9.3 Evaluation of ( 1Ip) and (2)~) Theorem
9.4 For every odd prime p we have
(lIPI
= (l)(p1)‘2
=
1
1 if p = 1 (mod 4), ifp ~ 3 (mod4)
PROOF. By Euler’s criterion we have ( 1 Ip) = ( l)(p r)12 (mod p). Since
each member of this congruence is 1 or  1 the two members are equal. 0 Theorem 9.5 For every odd prime p we have (4p)
= (1)@1)/8
=
1
1 f p = f 1 (mod 8), ifp= +3 (mod8).
PROOF. Consider the following 0,  1)/2 congruences: p  1 = l(  1)’
(mod P)
2 E 2(1)2
(mod P)
p  3 E 3(  1)3
(mod P)
4 E 4(
(mod P)
r E q
1)4
( l)(P IV2 (mod p),
where r is either p  (p  1)/2 or (p  1)/2. Multiply that each integer on the left is even. We obtain !( 1)1+2+,..+(~1)/2
these together and note
(mod
p).
This gives us 2(Pe’V2(!?$)!
G (!!)!(1)(pzm1)~8
(mod p).
181
9: Quadratic
residues and the quadratic
reciprocity
law
Since (0,  1)/2)! + 0 (mod p) this implies 2’P I)/2 = (_ I)@ I)/8 (mod p). By Euler’s criterion we have 2u ‘)I2 = (21~) (mod p), and since each member is 1 or  1 the two members are equal. This completes the proof. 0
9.4 Gauss’ lemma Although Euler’s criterion gives a straightforward method for computing (n Ip), the calculation may become prohibitive for large n since it requires raising n to the power (p  1)/2. Gauss found another criterion which involves a simpler calculation. Theorem 9.6 Gauss’ lemma. Assume n f 0 (mod p) and consider the least positive residues mod p of the following (p  1)/2 multiples of n: Pl n, 2n, 3n, . . , ~
2
n
If m denotes the number of these residues which exceed p/2, then
(nld = ( 1)“. PROOF.The numbers in (3) are incongruent mod p. We consider their least positive residues and distribute them into two disjoint sets A and B, according as the residues are p/2. Thus A = h, a2,. . .14J where each Ui 3 tn (mod p) for some t I (p  1)/2 and 0 < ai < p/2; and B = h b2, . . . , kl where each bi = sn (mod p) for some s I (p  1)/2 and p/2 < bi < p. Note that m + k = (p  1)/2 since A and B are disjoint. The number m of elements in B is pertinent in this theorem. Form a new set C of m elements by subtracting each bi from p. Thus C
=
{cl,
c2,.
. , cm},
where ci = p  bi.
Now 0 < ci < p/2 so the elements of C lie in the same interval as the elements of A. We show next that the sets A and C are disjoint. Assume that ci = uj for some pair i and j. Then p  bi = oj, or Uj + bi ~0 (mod p). Therefore tn + sn = (t + s)n = 0 (mod p)
for some s and t with 1 I t < p/2, 1 I s < p/2. But this is impossible since p $ n and 0 < s + t < p. Therefore A and C are disjoint, so their union 182
9.4: Gauss’ lemma
A u C contains m + k = (p  1)/2 integers in the interval [l, (p  1)/2]. Hence AuC=
{a,,a, )...) &,Cl,C2 )...) c,} =
Pl 1,2 )...) 7
i Now form the product of all the elements in A u C to obtain
( ) Pl, 2
ala2 . . . akclc2 . . . c, =
Since ci = p  bi this gives
..
US
Pl ( )
! = Cl142 . . . ah
2
.
 WP
 b2). .  (P  U
= ( l)mquz . . . akblba . . . b,
(mod P)
E ( l)“n(2n)(3n) . . .
(mod p) (mod P).
Canceling the factorial we obtain dp lu2 = ( 1)” (mod p). Euler’s criterion shows that (  1)” = (n 1p) (mod p) hence ( 1)” = (n 1p) and the proof of Gauss’ lemma is complete. 0 To use Gauss’ lemma in practice we need not know the exact value of m, but only its parity, that is, whether m is odd or even. The next theorem gives a relatively simple way to determine the parity of m. Ttieorem 9.7 Let m be the number defined in Gauss’ lemma. Then + (n  1) y
m=
In particular,
PROOF.
(mod 2).
ifn is odd we have
Recall that m is the number of least positive residues of the numbers n,2n,3n
,...,
Pl n
2 183
9: Quadratic
residues and the quadratic
reciprocity
law
which exceed p/2. Take a typical number, say tn, divide it by p and examine the size of the remainder. We have where0 < ip}E
;=[;]+{;],
< 1,
SO
say, where 0 < rt < p. The number I, = tn  p[tn/p] is the least positive residue of tn modulo p. Referring again to the sets A and B used in the proof of Gauss’ lemma we have { rl,r2,...,r(pl)i2~
= {ul,u2,...,uk,bl,...,b,}.
Recall also that
1,2,. . . , q
= {al, $7,. . . , &, cl,. . . , c,}
where each ci = p  bi. Now we compute the sums of the elements in these sets to obtain the two equations
and
In the first equation we replace flUi
r,
by its definition to obtain
+ j~~bj = n(P~‘t
 ~~‘[“]
r=1
P
t=1
The second equation is mp + iui i=l

i bj = (‘T’t. t=1
j=l
Adding this to the previous equation we get
[I =(n+1)fg;y2["]. (p1)/2
mp + 2 ~ i=l
Ui =
(fl +
l)@i)“t
t=1

p
C
r=1
r=1
184
tn
p
P
9.5 : The quadratic
reciprocity
law
Now we reduce this modulo 2, noting that n + 1 = n  1 (mod 2) and p E 1 (mod 2), and we obtain
m= (n 1)v
+ ‘py2[z] (mod2), t=1
P
which completes the proof.
0
9.5 The quadratic reciprocity law Both Euler’s criterion and Gauss’ lemma give straightforward though sometimes lengthy procedures for solving the first basic problem of the theory of quadratic residues. The second problem is much more difficult. Its solution depends on a remarkable theorem known as the quadratic reciprocity law, first stated in a complicated form by Euler in the period 17441746, and rediscovered in 1785 by Legendre who gave a partial proof. Gauss discovered the reciprocity law independently at the age of eighteen and a year later in 1796 gave the first complete proof. The quadratic reciprocity law states that if p and q are distinct odd primes, then (p 1q) = (q (p) unless p = q = 3 (mod 4), in which case (p 1q) =  (q 1p). The theorem is usually stated in the following symmetric form given by Legendre. Theorem 9.8 Quadratic reciprocity law. If p and q are distinct odd primes, then (4)
(jlq)(q1p)
= (1)(pl)(q1)/4.
PROOF. By Gauss’ lemma and Theorem 9.7 we have (4lP) = C1)”
where mr
Similarly,
@Id = C1) where
Hence (Plqklp)
= ( lJm+“, and (4) follows at once from the identity
185
9: Quadratic residuesand the quadratic reciprocity law
To prove (5) consider the function fk
Y) = w  PY.
If x and y are nonzero integers thenf(x, y) is a nonzero integer. Moreover, as x takes the values 1, 2, . . . , (p  1)/2 and y takes the values 1, 2, . . . , (q  1)/2 thenf(x, y) takes plq1 22
values, no two of which are equal since j(x, Y)  j(x’, Y’) = “ox  x’, y  Y’) z 0. Now we count the number of values off(x, y) which are positive and the number which are negative. For each fixed x we have f(x, y) > 0 if and only if y < qx/p, or y I [qx/p]. Hence the total number of positive values is
Similarly, the number of negative values is (q1)/2
4. =II1 py
y=l
Since the number of positive and negative values together is plq1 22
this proves (5) and hence (4).
0
Note. The reader may find it instructive to interpret the foregoing proof of (5) geometrically, using lattice points in the plane.
At least 150 proofs of the quadratic reciprocity law have been published. Gauss himself supplied no less than eight, including a version of the one just given. A short proof of the quadratic reciprocity law is described in an article by M. Gerstenhaber [25].
9.6 Applications
of the reciprocity
law
The following examples show how the quadratic reciprocity law can be used to solve the two basic types of problems in the theory of quadratic residues. EXAMPLE 1 Determine whether 219 is a quadratic residue or nonresidue
mod 383. 186
9.7: The Jacobi symbol
Solution We evaluate the Legendre symbol (2191383) by using the multiplicative property, the reciprocity law, periodicity, and the special values ( 1I p) and (2 1p) calculated earlier. Since 219 = 3 .73 the multiplicative property implies (219 1383) = (3 1383)(73 1383). Using the reciprocity law and periodicity we have (31383) = (38313)(1)‘3831)(31)/4 = (113) = (1)‘31)/2
= 1,
and (731383) = (383)73)(1)‘3831)(731)‘4 =
(_
l)K73F
1)/S
=
= (18173) = (2173)(9173)
1.
Hence (2191383) = 1 so 219 is a quadratic residue mod 383. EXAMPLE 2 Determine those odd primes p for which 3 is a quadratic residue
and those for which it is a nonresidue. Solution Again, by the reciprocity law we have (3lp) = (p,3)(l)‘p1~31)~4
= (1)‘Pypl3).
To determine (p 13)we need to know the value of p mod 3, and to determine l)(PI)/2 we need to know the value of (p  1)/2 mod 2, or the value of p mod 4. Hence we consider p mod 12. There are only four cases to consider, p E 1,5,7, or 11 (mod 12), the others being excluded since p is odd. Case 1. p = 1 (mod 12). In this case p G 1 (mod 3) so (pi 3) = (113) = 1. Also p E 1 (mod 4) so (p  1)/2 is even, hence (3 Ip) = 1. Case 2. p = 5 (mod 12). In this case p E 2 (mod 3) so (~13) = (213) = (4)‘32W3 = 1. A gain, (p 1)/2 is even since p = 1 (mod 4), so (3 Ip) = (_
Case 3. p E 7 (mod 12). In this case p = 1 (mod 3), so (pi 3) = (113) = 1. Also (p  1)/2 is odd since p E 3 (mod 4), hence (3 I p) =  1. Case 4. p G 11 (mod 12). In this case p = 2 (mod 3) so (pJ 3) = (2 13) =  1. Again (p  1)/2 is odd since p  3 (mod 4), hence (3 Ip) = 1. Summarizing the results of the four cases we find 3Rp if p E f 1 (mod 12) 3Rp ifp = +5 (mod 12).
9.7 The Jacobi symbol To determine if a composite number is a quadratic residue or nonresidue mod p it is necessary to consider several cases depending on the quadratic character of the factors. Some calculations can be simplified by using an extension of Legendre’s symbol introduced by Jacobi. 187
9: Quadratic residues and the quadratic reciprocity law
If P is a positive odd integer with prime factorization
Definition
the Jacobi symbol (n I P) is defined for all integers n by the equation (6)
tnlP)
=
fitnIP?’ i=l
where (n 1pi) is the Legendre symbol. We also define (n ) 1) = 1. The possible values of (n I P) are 1,  1, or 0, with (n(P) = 0 if and only if (n, P) > 1.
If the congruence x2 = n (mod P)
has a solution then (n Ipi) = 1 for each prime pi in (6) and hence (n 1P) = 1. However, the converse is not true since (n 1P) can be 1 if an even number of factors  1 appears in (6). The reader can verify that the following properties of the Jacobi symbol are easily deduced from properties of the Legendre symbol. Theorem 9.9 Zf P and Q are odd positive integers, we have
(4 (mlP)(nlP) = WlP), 09 (n IP)(n IQ) = b IPQ), (c) (m 1P) = (n I P) whenever m = n (mod P), (d) (a2n (P) = (n I P) whenever (a, P) = 1.
The special formulas for evaluating the Legendre symbols ( 1 Ip) and (2 1p) also hold for the Jacobi symbol. Theorem 9.10 If P is an odd positive integer we have (lip)
(7)
= (l)(P1)/2
and (ZIP) = (l)‘P’l)/f
(8)
PROOF. Write P = p1 p2 . . . pm where the prime factors pi are not necessarily distinct. This can also be written as P = ii(l i=l
188
+ pi  1) = 1 + ~ (pi  1) + C(pi  l)(pj  1) + “‘. i=l
i+j
9.7 : The Jacobi symbol
But each factor pi  1 is even so each sum after the first is divisible by 4. Hence Pcl+
F(Pil)(mod4), i=l
or i (P  1) E 5 i (pi  1) (mod 2). i=l
Therefore (lip)
= J+pi)
= fi(l)(Pfl)~2
i=l
= (1)(pl)‘2,
i=l
which proves (7). To prove (8) we write p2 = fi (1 + pi2  1) = 1 + f (pi’  1) + C(pi”  l)(Pj2  l) + ’ ’ * ’ i=l
i=l
i#j
Since pi is odd we have pi2  1 E 0 (mod 8) so p2 E 1 + f (pi2  1) (mod 64) i=l
hence $ (I’”  1) = f
f (pi2  1) (mod 8).
i=l
This also holds mod 2, hence (ZIP) = fi(2,p,) i=l
= f&L)(““1’1”
= (l)(P21)/8,
i=l
which proves (8).
q
Theorem 9.11 Reciprocity law for Jacobi symbols. IfP and Q are positive odd integers with (P, Q) = 1, then (P) Q)(Q (P) = ( l)“PROOF.
‘)(Q ‘)?
Write P = pl . . . pm, Q = q1 +. . q,,, where the pi and qi are primes.
Then (PIQ)(QIP)
= fi fI(Pi/qj)(qj/pJ i=l
= tl)‘,
j=l
say. Applying the quadratic reciprocity law to each factor we find that Y = ~ ~ ~ (pi  1) ~ (qj  1) = ~ f (pi  1) ~ ~ (qj  1). i=l j=l i=l j=l 189
9 : Quadratic
residues and the quadratic
reciprocity
law
In the proof of Theorem 9.10 we showed that itl k (pi  1) E i (P  1) (mod 2), and a corresponding congruence holds for 1 gqj  1). Therefore rTV
(mod2),
which completes the proof.
0
EXAMPLE 1 Determine whether 888 is a quadratic residue or nonresidue of
the prime 1999. Solution We have (88811999) = (411999)(211999)(111I 1999) = (11111999). To calculate (111 I 1999) using Legendre symbols we would write (11111999) = (3 ( 1999)(37 I 1999) and apply the quadratic reciprocity law to each factor on the right. The calculation is simpler with Jacobi symbols since we have (11111999) = (19991111) = (lllll)
= 1.
Therefore 888 is a quadratic nonresidue of 1999. EXAMPLE 2 Determine whether  104 is a quadratic residue or nonresidue of
the prime 997. Solution Since 104 = 2.4.13
we have
(1041997) = (l/997)(21997)(131997) = (997113)= (9113)=
= (131997) 1.
Therefore  104 is a quadratic nonresidue of 997.
9.8 Applications
to Diophantine
equations
Equations to be solved in integers are called Diophantine equations after Diophantus of Alexandria. An example is the equation (9)
y2 = x3 + k
where k is a given integer. The problem is to decide, for a given k, whether or not the equation has integer solutions x, y and, if so, to exhibit all of them. 190
9.8 : Applications to Diophantine equations
We discuss this equation here partly because it has a long history, going back to the seventeenth century, and partly because some cases can be treated with the help of quadratic residues. A general theorem states that the Diophantine equation Y2 = f(x) has at most a finite number of solutions iff(x) is a polynomial of degree 23 with integer coefficients and with distinct zeros. (See Theorem 418 in LeVeque [44], Vol. 2.) However, no method is known for determining the solutions (or even the number of solutions) except for very special cases. The next theorem describes an infinite set of values of k for which (9) has no solutions. Theorem 9.12 The Diophantine
equation y2 = x3 + k
(10)
has no solutions ifk has the form k = (4n  1)3  4m2,
(11)
where m and n are integers such that no prime p G  1 (mod 4) divides m. PROOF. We assume a solution x, y exists and obtain a contradiction considering the equation modulo 4. Since k =  1 (mod 4) we have y2 E x3
(12)
by
 1 (mod 4).
Now y2 E 0 or 1 (mod 4) for every y, so (12) cannot be satisfied if x is even or if x =  1 (mod 4). Therefore we must have x E 1 (mod 4). Now let a=4n1 so that k = a3  4m2, and write (10) in the form (13)
y2 + 4m2 = x3 + a3 = (x + a)(x2  ax + a’).
Since x = 1 (mod 4) and a =  1 (mod 4) we have (14)
X2
ax+a2la+a2
1 (mod4).
Hence x2  ax + a2 is odd, and (14) shows that all its prime factors cannot be = 1 (mod 4). Therefore some prime p E  1 (mod 4) divides x2  ax + a2, and (13) shows that this also divides y2 + 4m2. In other words, (15)
y2  4m’
(mod p) for some p E  1 (mod 4).
But p $ m by hypothesis, so (  4m2 1p) = ( 11p) =  1, contradicting (15). This proves that the Diophantine equation (10) has no solutions when k has 0 the form (11). 191
9 : Quadratic residuesand the quadratic reciprocity law
The following table gives some values of
k
covered by Theorem 9.12.
n
0
0
0
011
112222
m
1
2
4
512
4
k
5
17
100
65
5
23 11 37
73
12
4
5
339 327 279 243
Note. All solutions of (10) have been calculated when k is in the interval  100 I k I 100. (See reference [32].) No solutions exist for the following positive values of k I 100: k = 6,7,
11, 13, 14,20,21,23,29,32,34,39,42,45,46,47,
51,53, 58,
59,60,61,62,66,67,69,70,74,75,77,78,83,84,85,86,87,88,90, 93,95, 96.
9.9 Gauss sums and the quadratic reciprocity law This section gives another proof of the quadratic reciprocity law with the help of the Gauss sums G(n, x) =
(16)
c X(r)e2ninr’p, rmodp
where X(r) = (rip) is the quadratic character mod p. Since the modulus is prime, x is a primitive character and we have the separability property (17)
Gh xl = (n IdGU,
xl
for every n. Also, Theorem 8.11 implies that 1G( 1, x) 1’ = p. The next theorem shows that G(1, x)2 is fp. Theorem 9.13 !fp is an odd
prime
and x(r) = (r 1p) we have
W, xl2 = ( 1 IPIP.
(18)
PROOF.We have pl
G(l, x)’ = 1 r=l
pl
c
(rIp)(sIp)e2ni(r+S)‘p.
s=l
For each pair r, s there is a unique t mod
p
such that s = tr (mod
(~IPM~IP)= (rlp)@rlp) = (r21pNlp) = @IPI. Hem pl
G(l, x)2 = 1 1=1
pl
p1
p1
1 (tIp)e2zir(1+f)lp = r=l
The last sum on r is a geometric sum given by p1 Znir(1 Ce
r=l 192
+
O/p = ip
1 ifp#(l + t),  1 ifpl(1 + t).
p),
and
9.9: Gauss sums and the quadratic reciprocity law
Therefore p2 W,X)‘=

p1
C(tlp)+(pt=1
l)(p
l/p)=

C(t~p)+~(ll~) i=l
= (llPhJ Cl
since Ecri (t lp) = 0. This proves (18).
Equation (18) shows that G(l, x)2 is an integer, so G(l, x)“ ’ is also an integer for every odd q. The next theorem shows that the quadratic reciprocity law is connected to the value of this integer modulo q. Theorem 9.14 Let p and q be distinct odd primes and let x be the quadratic character mod p. Then the quadratic reciprocity law (qlp) = (l)(pl)(qyplq)
(19) is equivalent
to the congruence
GO, xJqwl = (qlp) (mod 4).
(20) PROOF.
(21)
From (18) we have G(l,
.#
1 = (_ 1 Ip)'"
1)/2p(q 1)/z =
(_
l)(P
I)@
1)/4#q
1)/z.
By Euler’s criterion we have p’q 1)‘2 = (p 1q) (mod q) so (21) implies (22)
G(1, x)4l z (1)(P1)(q1)/4(plq)
(mod q).
If (20) holds we obtain (qlp) = (l)(p1)(q1)‘4(plq)
(mod q)
which implies (19) since both members are &1. Conversely, if (19) holds then (22) implies (20). Cl The next theorem gives an identity which we will use to deduce (20). Theorem9.15 Zfp and q are distinct odd primes and ifx is the quadratic character mod p we have
(23)
G(Lx)~~ = (qlp)
1
... I zddp(T1... r,Id.
r,modp
rl++rq=q:modp,
PROOF. The Gauss sum G(n, 2) is a periodic function of n with period p. The same is true of G(n, x)” so we have a finite Fourier expansion
G(n, x)” =
c
aq(m)e2”imn’P,
mmodp
193
9 : Quadratic
residues and the quadratic
reciprocity
law
where the coefficients are given by a,(m) = f
(24)
1 G(n, X)qe2nimn’p. nmodp
From the definition of G(n, x) we have G(n, x)’ =
1 II
=
(rl Ip)e2ninr1’p . .  , ~dp(T,Ipk2xi”rq’p
modp
zdp(Tl
r,gdp...,
. ..r.lp)e2~in(rl+...+r~~,p,
4
so (24) becomes a,(m) = L
. . . , gdFl
1
P r,modp
. . . rqlp)
4
1
e2nin(rl+...+lqm)ip.
nmodp
The sum on n is a geometric sum which vanishes unless
rI + . . . + rq E
m (mod p), in which case the sum is equal to p. Hence (25)
Now we return to (24) and obtain an alternate expression for u,(m). Using the separability of G(n, x) and the relation (n 1~)~ = (n Ip) for odd q we find a,(m) = i G(l, x)’ c
(nlp)e2Rimn’P = $ G(l, X)qG(m,
x)
nmodp
= k W, x)q(mIp)G( 1, x1 = (mIpN31, xIqml since
W, x)G( 4 x) = G(4 x)G(l, xl = IW, x)1’ =
P.
In other words, G(l, x)“ ’ = (m Ip)u,(m). Taking m = q and using (25) we obtain (23). 0 tioo~
OF THE RECIPROCITY
LAW.
To deduce the quadratic reciprocity law
from (23) it suffices to show that (26)
c
. ..
rtmodp
r
q;dp@’
. ..rqlp)
=
1
(modq),
where the summation indices rr, . . . , rq are subject to the restriction (27) 194
r1
+
. . . + r4 E q (mod p).
9.10: The reciprocity law for quadratic Gauss sums
If all the indices yl, . . . , rq are congruent to each other mod p, then their sum is congruent to qrj for each j = 1,2, . . . , q, so (27) holds if, and only if, qrj E 4 (mod P), that is, if, and only if rj = 1 (mod p) for each j. In this case the corresponding summand in (26) is (1 lp) = 1. For all other choices of indices satisfying (27) there must be at least two incongruent indices among rl, . . . , Ye. Therefore every cyclic permutation of rl, . . . , r4 gives a new solution of (27) which contributes the same summand, (rl f . . r4 1p). Therefore each such summand appears q times and contributes 0 modulo q to the sum. Hence the only contribution to the sum in (26) which is nonzero modulo q is (1 Ip) = 1. This completes the proof. q
9.10 The reciprocity sums
law for quadratic
Gauss
This section describes another proof of the quadratic reciprocity law based on the quadratic Gauss sums G(,,;
m)
=
f
e2ninr2/m.
r=l
If p is an odd prime and p $ II we have the formula (29) G(n; P) = (a IPM ; P) which reduces the study of the sums G(n; p) to the case n = 1. Equation (29) follows easily from (28) or by noting that G(n; p) = G(n, x), where x(n) = (n 1p), and observing that G(n, x) is separable. Although each term of the sum G(l; p) has absolute value 1, the sum itself has absolute value 0, & or A. In fact, Gauss proved the remarkable formula CJm (30) G(l; m) = i ,,/%(l + i)(l + ePni’“12)=
’ ‘J ;I +” i)&
ifm ifm ifm if m
= 1 (mod4) = 2 (mod4) = 3 (mod4) E 0 (mod 4)
for every m 2 1. A number of different proofs of (30) are known. We will deduce (30) by treating a related sum ml
S(a, m) = 1 eniar2’m, r=O
where a and m are positive integers. If a = 2, then S(2, m) = G(l; m). The sums S(a, m) enjoy a reciprocity law (stated below in Theorem 9.16) which implies Gauss’ formula (30) and also leads to another proof of the quadratic reciprocity law. 195
9 : Quadratic
residues and the quadratic
reciprocity
law
Theorem 9.16 If the product ma is even, we have m l+i a S(m, 4, J( Jz ) where the bar denotes the complex conjugate. S(a, m) =
(31)
Note. To deduce Gauss’ formula (30) we take a = 2 in (31) and observe that S(m, 2) = 1 + emnimi2. This proof is based on residue calculus. Let g be the function defined by the equation
PROOF.
g(z)
(32)
=
ml C
enia(~+*)2/me
r=O
Then g is analytic everywhere, and g(0) = S(a, m). Since ma is even we find a1 g(z
+
1) _ g(z)
=
eniazzlm(e*niaz
_
1) =
eniaz2/m(e2rriz
_
1) 2
e2ninz.
n=O
Now definefby
the equation f(z)
= .,:!y
1.
Then f is analytic everywhere except for a firstorder pole at each integer, andfsatisfies the equation (33) where (34)
f(z + 1) = f(z) + cp(z), cp(z)
=
eniaz2/m
al 1
e2ninz.
n=O
The function cpis analytic everywhere. At z = 0 the residue off is g(O)/(2ri) and hence S(a, m) = g(0) = 2ni Res f(z) = f(z) dz, sY z=o where y is any positively oriented simple closed path whose graph contains only the pole z = 0 in its interior region. We will choose y so that it describes a parallelogram with vertices A, A + 1, B + 1, B where
(35)
A =  i  Reni14and B =  z1 + Reni14, as shown in Figure 9.1. Integratingfalong
196
y we have
9.10:
The reciprocity
law for quadratic
Gauss sums
Figure 9.1
In the integral JiI : f we make the change of variable w = z + 1 and then use (33) to get l3+1
sA+1
f(w)dw
=
J)tz
+
l)dz
=
J;f(z)dz
+
J)le)dz.
Therefore (35) becomes B
(36)
S(a, m) =
IA
A+1
d.4 dz + A
B+l
f(z) dz 
I
B
J(z) dz.
I
Now we show that the integrals along the horizontal segments from A to A + 1 and from B to B + 1 tend to 0 as R + + co. To do this we estimate the integrand on these segments. We write (37) and estimate the numerator and denominator separately. On the segment joining B to B + 1 we let y(t) = t + Re”i’4,
where it
1 2
I.
1 2
From (32) we find ml
(38)
Ig[y(t)] 1 < C exp nia@ + Rini’4 + r)2 , r=O I II i
where exp z = e’. The expression in braces has real part na(@R
+ R2 + $rR) m 197
9 : Quadratic
residues and the quadratic
reciprocity
law
Since )eX+iY1= eXand exp{  ~~a,/%R/rn} I 1, each term in (38) has absolute value not exceeding exp{  naR’/m}exp{ ,,hrcatR/m). But  l/2 I t I l/2, so we obtain the estimate 1s[r(t)] 1 5 me afiWWOe  naR’/m For the denominator in (37) we use the triangle inequality in the form le2niz  1 ) 2 1leZniz) 11, Since (exp{2rciy(t)} 1= exp{  2nR sin(z/4)} = exp{ $&CR},
we find
le2WW  11 2 1  eJ2nR. Therefore on the line segment joining B to B + 1 we have the estimate men+‘%R/(2m)
If(
5
noR*/m
1 _ ,:.R
= o(l)
as R + +co.
A similar argument shows that the integrand tends to 0 on the segment joining A to A + 1 as R + + co. Since the length of the path of integration is 1 in each case, this shows that the second and third integrals on the right of (36) tend to 0 as R + + co. Therefore we can write (36) in the form B q(z) dz + o(l) as R + +co. S(a, m) = (39) sA To deal with the integral j: cp we apply Cauchy’s theorem, integrating cparound the parallelogram with vertices A, B, IX, LX, where M = B + $ = Renii4, (SeeFigure 9.2.) Since cpis analytic everywhere, its integral around this parallelogram is 0, so
Figure 9.2
198
9.10:
The reciprocity law for quadratic Gausssums
in (34) an argument similar to that Because of the exponential factor enioz2/m given above shows that the integral of cpalong each horizontal segment + 0 as R + + co. Therefore (40) gives us q + o(l)
asR+
+co,
and (39) becomes a S(a, m) =
(41)
q(z)dz + o(l) s (1
as R + +cq
where u = Reni14. Using (34) we find J)(z)
dz = 1:; ~~aeni‘~‘“e2~i”z dz = ~~~enimn2~aZ(a, m, n, R),
where Z(a,m,n,R)=
,:expr;(z+y)z]dz.
Applying Cauchy’s theorem again to the parallelogram with vertices  01,a, u  (WI/~), and c1  (WI/U), we find as before that the integrals along the horizontal segments + 0 as R + + co, so Z(o,m,n,R)=~~~~~~,~p{~(z+~)2jdz+o(1) The change of variable w = &&r(z
asR+
+co.
+ (n&z)) puts this into the form as R , + co.
Z(a, m, n, R) =
Letting R + + co in (41), we find a1
(42)
eniw*
S(u, m) = 1 enimn2’a
By writing T = fiR,
dw.
n=O
we see that the last limit is equal to Ten’/4 lim eniW2 dw = 1 s  T@T’,‘l T++m
say, where Z is a number independent of a and m. Therefore (42) gives us S(a, m) =
F ZS(m, a), ll
To evaluate I we take a = 1 and m = 2 in (43). Then S(l, 2) = 1 + i and S(2, 1) = 1, so (43) implies I = (1 + i)/a, and (43) reduces to (31). 0 199
9 : Quadratic
residues and the quadratic
reciprocity
law
Theorem 9.16 implies a reciprocity law for quadratic Gauss sums. Theorem 9.17 Zfh > 0, k > 0, h odd, then i 7
G(h; k) =
(44)
(1 + enihklZ)G(k; h).
J
PROOF. Take a = 2h, m = k in Theorem 9.16 to obtain (45)
G(h; k) = S(2h, k) =
We split the sum on r into two parts corresponding to even and odd r. Forevenrwewriter = 2swheres = 0,1,2,...,h  l.Foroddrwenotethat (r + 2h)2 = r2 (mod 4h) so the sum can be extended over the odd numbers in any complete residue system mod 2h. We sum over the odd numbers in the interval h I r < 3h, writing r = 2s + h, where s = 0, 1,2, . . . , h  1. (The numbers 2s + h are odd and distinct mod 2h.) This gives us Zh,zo
1 e
nikr2/(2h)
=
hl s~oe~ik(2~)‘i(2h)
=
hl sgoe2dcs2/h(l
=(l+e
+
+
hl s;oenik(2s+h)2/(2hl
,dW2)
 zihk’2)G(k; h).
cl
Using this in (45) we obtain (44).
9.11 Another proof of the quadratic reciprocity law Gauss’ formula (30) leads to a quick proof of the quadratic reciprocity law. First we note that (30) implies G(l; k) = i(k
1j2/4&
if k is odd. Also, we have the multiplicative G(m;n)G(n;m) = G(l;mn)
property (see Exercise 8.16(a)) if(m,n) = 1.
Therefore, if p and q are distinct odd primes we have
(3.~;q) = @Iq)W; 4 = (plq)i’q1)2’4,h G(q; P) = (qlp)G(l;
P) = (q.lp)i(p1)2’4&
and G(p; q)G(q; p) = G(l; pq) = i(pq1)2’4&. 200
Exercises for Chapter 9
Comparing the last equation with the previous two we find (p14)(41p)il(41)2+(P1)*t/4
=
i(P41)2/4,
and the quadratic reciprocity law follows by observing that i (@4
1)2(9
1)2(P
1)2)/4
=
(_
l)(P
l)(9
1)/4e
Cl
Exercises for Chapter 9 1. Determine those odd primes p for which ( 3 Ip) = 1 and those for which ( 3 Ip) = 1. 2. Prove that 5 is a quadratic residue of an odd prime p if p = f 1 (mod lo), and that 5 is a nonresidue if p = + 3 (mod 10). 3. Let p be an odd prime. Assume that the set { 1,2, . . . , p  1) can be expressed as the union of two nonempty subsets S and ?; S # T, such that the product (mod p) of any two elements in the same subset lies in S, whereas the product (mod p) of any element in S with any element in T lies in T. Prove that S consists of the quadratic residues and T of the nonresidues mod p. 4. Letf(x) be a polynomial which takes integer values when x is an integer. (a) If a and b are integers, prove that ,~dp(f(~x
+ b)lp) = x~dp(f(41~)
if@, P) = 4
and that x~ddp(d’(x)I~)
= (~IP)~~~UWIP)
for all a.
(b) Prove that c (ax+blp)=O
if(a,p)=l.
xmodp
(c) Let f(x) = x(ax + b), where (a, p) = (b, p) = 1. Prove that p1
p1
x;IW)l~)
= ,TI(a + bxlp) = (alp).
[Hint: As x runs through a reduced residue system mod p, so does x’, the reciprocal of x mod p.] 5. Let a and jI be integers whose possible values are &1. Let N(a, /I) denote the number of integers x among 1,2,. . . , p  2 such that
blp) = a
and(x+
lip)=/?,
where p is an odd prime. Prove that p2
Wa, 8) = C (1 + a(xIp)I{l + Rx + 1IpI}, x=1
201
9: Quadratic
residues and the quadratic
reciprocity
law
and use Exercise 4 to deduce that 4N(a,p) In particular
= p  2  /I  a/?  a(lip).
this gives N(l
l)=P4(llP) 4
’
N(1, _,)=.(,,l)=p2+4(1’p), N(l,
1)
= 1 I N&l).
6. Use Exercise 5 to show that for every prime p there exist integers x and y such that x2 + y2 + 1 = 0 (mod p). 7. Let p be an odd prime.
Prove each of the following
statements:
p1
(a) 1 r(r(p) = 0 r=l
ifp = 1 (mod 4).
p1
(b)
1 ,=l HP)
r=F
ifp=l
= 1 p1
p1
(c)
(mod4).
ifp=3
Cr’(rlp)=pCr(rIp)
r=1
(mod4.
,=1
p1
(d) 1 r3(rlp) r=l p1
= i pp~‘rz(rlp) I 1 P
if p = 1 (mod 4).
1
p1
(e) Cr4(rIp)=2pCr3(r.Ip)p* r=1 r=l [Hint:
Cr’(rlp) r=l
p  r runs through
the numbers
ifp= 1,2,
3 (mod4). , p  1 with r.]
8. Let p be an odd prime, p = 3 (mod 4), and let q = (p  1)/2. (a) Prove that
(1 
irkId = PT
~@IP)I
r=l
i (r(p). ,=1
[Hint: As r runs through the numbers 1,2, . . . , q then r and p  r together through the numbers 1,2, . . , p  1, as do 2r and p  2r.l (b) Prove that p1
((21~)
202

21 1 rklp) r=1
= p E (rip). ,= 1
run
Exercises for Chapter
9
9. If p is an odd prime, let x(n) = (nip). Prove that the Gauss sum G(n, x) associated with x is the same as the quadratic Gauss sum G(n; p) introduced in Exercise 8.16 if (n, p) = 1. In other words, if p J’ n we have G(n,
x) =
1 mmodp
~(+~~~~“‘p = i ezninr2/P= G(n; p). r=l
It should be noted that G(n, x) # G(n; p) if pin because G(p, x) = 0 but G(p; p) = p. 10. Evaluate the quadratic Gauss sum G(2; p) using one of the reciprocity laws. Compare the result with the formula G(2; p) = (2lp)G(l; p) and deduce that (21~) = ( l)@ i)18 if p is an odd prime.
203
10
Primitive Roots
10.1 The exponent of a number Primitive roots
mod m.
Let a and m be relatively prime integers, with m 2 1, and consider all the positive powers of a: a, a=, a3, . . .
We know, from the EulerFermat theorem, that aa = 1 (mod m). However, there may be an earlier power af such that as = 1 (mod m). We are interested in the smallest positivefwith this property. Definition
The smallest positive integer f such that af = 1 (mod m)
is called the exponent of a modulo m, and is denoted by writing f = exp,(a). If exp,(a) = q(m) then a is called a primitive root mod m. The EulerFermat theorem tells us that exp,(a) 5 q(m). The next theorem shows that exp,(a) divides q(m). Theorem
10.1 Given m 2 1, (a, m) = 1, let f = exp,(a). Then we have:
(a) ak = ah (mod m) if, and only if, k = h (mod f). (b) ak = 1 (mod m) if, and only if, k E 0 (modf). In particular,f (c) The numbers 1, a, a=, . . . , af  1 are incongruent mod m. 204
I q(m).
10.2 : Primitive
roots and reduced residue systems
hOOF. Parts (b) and (c) follow at once from (a), so we need only prove (a). If ak E ah (mod m) then ukVh E 1 (mod m). Write
k  h = qf + r,
where 0 I r < fi
Then 1 E akh = @+r
=  a’ (mod m), so r = 0 and k E h (mod f ). Conversely, if k E h (mod f) then k  h = qf so ukh E 1 (mod m) and hence uk  uh (mod m). cl
10.2 Primitive roots and reduced residue systems Theorem 10.2 Let (a, m) = 1. Then a is a primitive the numbers (1)
root mod m if, and only if,
a, u2, . . . ) uq(m)
form a reduced residue system mod m.
If a is a primitive root the numbers in (1) are incongruent mod m, by Theorem 10.1(c). Since there are q(m) such numbers they form a reduced residue system mod m. Conversely, if the numbers in (1) form a reduced residue system, then LX”(~)= 1 (mod m) but no smaller power is congruent to 1, so a is a primitive root. 0
PROOF.
Note. In Chapter 6 we found that the reduced residue classes mod m form a group. If m has a primitive root a, Theorem 10.2 shows that this group is the cyclic group generated by the residue class a^.
The importance of primitive roots is explained by Theorem 10.2. If m has a primitive root then each reduced residue system mod m can be expressed as a geometric progression. This gives a powerful tool that can be used in problems involving reduced residue systems. Unfortunately, not all moduli have primitive roots. In the next few sections we will prove that primitive roots exist only for the following moduli: m = 1, 2,4, pa, and 2p”,
where p is an odd prime and a 2 1. The first three casesare easily settled. The case m = 1 is trivial. Form = 2 the number 1 is a primitive root. For m = 4 we have (p(4) = 2 and 32 1 (mod 4), so 3 is a primitive root. Next we show that there are no primitive roots mod 2” if a 2 3. 205
10 : Primitive
roots
10.3 The nonexistence of primitive roots mod 2” for a 2 3 Theorem 10.3 Let x be an odd integer. Zfu 2 3 we have
x’(*“)‘* E 1 (mod 2”),
(4 so there are no primitive
root:; mod 2”.
PROOF. If CI= 3 congruence (2) states that x2 E 1 (mod 8) for x odd. This is
easily verified by testing x = 1, 3, 5, 7 or by noting that (2k + l)* = 4k’! + 4k + 1 = 4k(k + 1) + 1 and observing that k(k + 1) is even. Now we prove the theorem by induction on LXWe assume (2) holds for u and prove that it also holds for u + 1. The induction hypothesis is that xrpc2=)/2
= 1+
23,
where t is an integer. Squaring both sides we obtain x’J’c2”)= 1 + 2”
+ ‘t + 22at2 G 1 (mod
20L+‘)
because2cr2 IX + 1.This completes the proof since (~(2”) = 2” ’ = (p(2”+ ‘)/2. cl
10.4 The existence of primitive roots mod for odd primes p
p
First we prove the following lemma. Lemma 1 Given (a, m) = 1, let f’ = exp,(a). Then expmbk)
In particular, PROOF.
=
exp&) (k, f)
.
exp,(ak) = exp,(a) if, and only if, (k, f) = 1.
The exponent of ak is the smallest positive x such that axk s 1 (mod m).
This is also the smallest x > 0 such that kx = 0 (mod f). But this latter congruence is equivalent to the congruence
where d = (k, f). The smallest positive solution of this congruence is f/d, so exp,(ak) = f/d, as asserted. 0 206
10.4: The existenceof primitive roots mod p for odd primesp
Lemma 1 will be used to prove the existence of primitive roots for prime moduli. In fact, we shall determine the exact number of primitive roots mod p. Theorem 10.4 Let p be an odd prime and let d be any positive divisor of p  1. Then in every reduced residue system mod p there are exactly q(d) numbers a such that
exp,(a) = d. In particular, when d = q(p) = p  1 there are exactly cp(p  1) primitive roots mod p. PROOF.
We use the method employed in Chapter 2 to prove the relation
The numbers 1,2,. . . , p  1 are distributed into disjoint sets A(d), each set corresponding to a divisor d of p  1. Here we define A(d) = {x : 1 I x I p  1 and exp,(x) = d}.
Let f(d) be the number of elements in A(d). Thenf(d) 2 0 for each d. Our goal is to prove thatf(d) = q(d). Since the sets A(d) are disjoint and since each x = 1,2, . . . , p  1 falls into some A(d), we have 1 f(d) dip 1
= P  1.
But we also have 1 cp(4 = P  1 % 1
so
d,,clM4  S(d)} = 0. To show each term in this sum is zero it suffices to prove thatf(d) I q(d). We do this by showing that eitherf(d) = 0 orf(d) = q(d); or, in other words, thatf(d) # 0 impliesf(d) = q(d). Suppose that f(d) # 0. Then A(d) is nonempty so a E A(d) for some a. Therefore exp,(a) = d, hence ad  1 (mod p). But every power of a satisfies the same congruence, so the d numbers (3)
a, a2,. . . , ad
207
10: Primitive roots
are solutions
of the polynomial congruence xd  1  0 (mod p),
(4)
these solutions being incongruent mod p since d = exp,(a). But (4) has at most d solutions since the mod.ulus is prime, so the d numbers in (3) must be all the solutions of (4). Hence each number in A(d) must be of the form uk forsomek = 1,2,..., d. When is exp,(ak) = d? According to Lemma 1 this occurs if, and only if, (k, d) = 1. In other words, among the d numbers in (3) there are cp(d) which have exponent d modulo p. Thus we have shown that f(d) = q(d) iff(d) # 0. As noted earlier, this completes the proof. Cl
10.5 Primitive roots and quadratic residues Theorem 10.5 Let g be a primitive
root mod p, where p is an odd prime. Then
the even powers g2,g4,...,gpl are the quadratic
residues mod p, and the odd powers 9, g3, * . ., gp 2
are the quadratic PMoF.
nonresidues mod p.
If n is even, say n = 2m, then g” = (g”)’ so g” E x2 (mod p), where x = gm.
Hence g”Rp. But there are (p  1)/2 distinct even powers g2, . . . , gp’ modulo p and the same number of quadratic residues mod p. Therefore the even powers are the quadratic residues and the odd powers are the nonresidues. q
10.6 The existence of primitive roots mod
p”
We turn next to the case m = pa, where p is an odd prime and u 2 2. In seeking primitive roots mod pa it is natural to consider as candidates the primitive roots mod p. Let g be such a primitive root and let us ask whether g might also be a primitive root mod p2. Now gpml = 1 (mod p) and, since cp(p’) = p(p  1) > p  1, this g will certainly not be a primitive root mod p2 if gp 1 = 1 (mod p2). Therefore the relation gp.l f 1 (mod p’) is a necessary condition for a primitive root g mod p to also be a primitive root mod p2. Remarkably enough, this condition is also sujficient for g to be a primitive root mod p2 and, more generally, mod pa for all powers c12 2. In fact, we have the following theorem. 208
10.6: The existence of primitive
roots mod p”
Theorem 10.6 Let p be an odd prime. Then we have: (a) Zf g is a primitive root mod p then g is also a primitive all c( 2 1 if, and only if,
root mod pafor
9p ’ f 1 (mod p2).
(5)
(b) There is at least one primitive there exists at least one primitive
root g mod p which satisfies (5), hence root mod pa if ct 2 2.
We prove(b) first. Let g be a primitive root mod p. IfgP ’ f 1 (mod p2) there is nothing to prove. However, if gpr E 1 (mod p2) we can show that gr = g + p, which is another primitive root modulo p, satisfies the condition PROOF.
91
p ’ + 1 (mod p’).
In fact, we have 91p1 = (g + py1
= gp1 + (p  l)gpZp
+ tp2
=  gpl
+ (p2  p)gpm2 (mod p2)  1  pgpm2 (mod p’).
But we cannot have pgpW2 = 0 (mod p2) for this would imply gpe2 = 0 (mod p), contradicting the fact that g is a primitive root mod p. Hence 1 (mod p’), so (b) is proved. 91 p ’ f Now we prove (a). Let g be a primitive root modulo p. If this g is a primitive root mod p” for all a 2 1 then, in particular, it is a primitive root mod p2 and, as we have already noted, this implies (5). Now we prove the converse statement. Suppose that g is a primitive root mod p which satisfies (5). We must show that g is also a primitive root mod p” for all u 2 2. Let t be the exponent of g modulo p”. We wish to show that t = cp(p”).Since g’ = 1 (mod p”) we also have g1 E 1 (mod p) so cp(p)I t and we can write
t = w(P).
(6)
Now t I cp(p”)so q(p(p) 1cp(p”).But cp(p”) = pa ‘(p  1) hence 4(P  l)lP”%
 1)
which means q 1pa ‘. Therefore q = pB where b I a  1, and (6) becomes t = ps(p  1).
If we prove that B = c1 1 then t = cp(p”)and the proof will be complete. Suppose, on the contrary, that j? < c1 1. Then /I I a  2 and we have t = pQ
 l)lp”2(p
 1) = cp(p”‘).
Thus, since cp(p” ‘) is a multiple of t, this implies, (7)
ga(P” ‘) E 1
(mod
pa).
209
10: Primitive
roots
But now we make use of the following Lemma which shows that (7) is a contradiction. This contradict:ion will complete the proof of Theorem 10.6. cl Lemma 2 Let g be a primitive
root modulo p such that 9 p1 f 1 (mod p’).
(8)
Then for every a 2 2 we have g’pl:P”
(9)
f 1 (mod p”).
 ‘)
PROOF OF LEMMA 2. We use induction on a. For a = 2, relation (9) reduces to (8). Suppose then, that (9) holds for a. By the EulerFermat theorem we have
9‘@(Pa‘)
E
1
(mod
p=‘)
so
gv,(,P=')
=
1 +
kp"l
where p ,j’ k because of (9). Raising both sides of this last relation to the pth power we find gVP(P=‘)
=
(1
+
kp=l)P
=
1
+
kp=
+
k2
p(p;
‘)
pz(=l)
+
rp3(=1).
Now 2a  1 2 a + 1 and 3a  3 2 a + 1 since a 2 2. Hence, the last equation gives us the congruence gq(p”) e 1 + kp” (mod pa+ ‘) where p J’ k. In other words, gVp(P’)f 1 (mod p”+ ‘) so (9) holds for a + 1 if it holds for a. This completes the proof of Lemma 2 and also of Theorem 10.6. 0
10.7 The existence of primitive roots mod 2p” Theorem 10.7 Zf p is an odd prime and a 2 1 there exist odd primitive g modulo p”. Each such g is also a primitive
roots
root modulo 2~“.
PROOF. If g is a primitive root modulo pa so is g + pa. But one of g or g + pa is odd so odd primitive roots mod pa always exist. Let g be an odd primitive root mod pa and let f be the exponent of g mod 2~“. We wish to show that
f = (~(2~7. Now f Iv(~P”),
and 442~“) = cp(2)cpW) = cp(P”) so f I cp(P”). On
the other hand, g/ = 1 (mod 2~“) so gf = 1 (mod p”), hence cp(p”)lf since g is a primitive root mod pa. Thereforef = cp(p”)= (~(2~7, so g is a primitive root mod 2~“. Cl 210
10.8: The nonexistence
10.8 The nonexistence of primitive the remaining cases
of primitive
roots in the remaining
cases
roots in
Theorem 10.8 Given m 2 1 where m is not of the form m = 1,2,4, pa, or 2p”, where p is an odd prime. Then for any a with (a, m) = 1 we have a+‘(m)‘2  1 (mod m), so there are no primitive
roots mod m.
PROOF.We have already shown that there are no primitive roots mod 2” if CI2 3. Therefore we can suppose that m has the factorization m = yplal
. . . psdp
where the pi are odd primes, s 2 1, and CI2 0. Since m is not of the form 1,2,4,p”or2pawehavecr22ifs=1ands22ifcr=Oor1.Notethat cp(m)= cp(W~h”‘)
. . . cp(P,“9.
Now let a be any integer relatively prime to m. We wish to prove that a’P(m)‘2 E 1 (mod m).
Let g be a primitive root mod pIa’ and choose k so that a E g“ (mod pIal).
Then we have (10) where
arpw2 = Wm)/2 E 9 f4’(Pla’) (mod plW) 9
t = kq(T)cp(p,““)
. . . (p(p,““)/2.
We will show that t is an integer. If c( 2 2 the factor (~(2’) is even and hence t is an integer. If a = 0 or 1 then s 2 2 and the factor cp(p2”‘)is even, so t is an integer in this case as well. Hence congruence (10) gives us arp(m)i2 E 1 (mod plbl).
In the same way we find (11)
a@‘(m)/2s 1 (mod piai)
foreachi = 1,2,... , s. Now we show that this congruence also holds mod 2”. If o! 2 3 the condition (a, m) = 1 requires a to be odd and we may apply Theorem 10.3 to write a+‘(2’)‘2E 1 (mod 2”). 211
10: Primitive
roots
Since (~(2”)1q(m) this gives us &‘(‘“)lz E 1 (mod 2’)
(12)
for a 2 3. Ifa I 2wehave cP’(~“)= 1 (mod 2”).
(13)
But s 2 1 so q(m) = rp(2”)4$pl”) . +. cp(p,“‘) = 2rq(2”) where Y is an integer. Hence q(2’) 1(p(m)/2 and (13) implies (12) for a I 2. Hence (12) holds for all c(. Multiplying together the congruences (11) and (12) we obtain LP(‘“)‘~  1 (mod m),
0
and this shows that a cannot be a primitive root mod m.
10.9 The number
of primitive
roots mod m
We have shown that an integer m 2 1 has a primitive root if and only if m == 1,2,4,p”or2p”,
where p is an odd prime and s12 1. The next theorem tells us how many primitive roots exist for each such m. Theorem 10.9 Zf m has a primitive root g then m has exactly cp(q(m)) incongruent primitive
roots and they are given by the numbers in the set
S= {g”:l
1. root mod p if p is a prime of the form 4q + 1, where q
5. Let m > 2 be an integer having a primitive root, and let (a, m) = 1. We write aRm if there exists an x such that a = x2 (mod m). Prove that: (a) aRm if, and only if, arp("')'* E 1 (mod m). (b) If aRm the congruence x2 = a (mod m) has exactly two solutions. (c) There are exactly q(m)/2 integers a, incongruent mod m, such that (a, m) = 1 and aRm. 6. Assume m > 2, (a, m) = 1, aRm. Prove that the congruence exactly two solutions if, and only if, m has a primitive root.
x2 = a (mod m) has
7. Let S,(p) = CnZ: k”, where p is an odd prime and n > 1. Prove that 0 (modp) UP)
=
{  1 (mod p)
1 8. Prove that the sum of the primitive
ifnf0
(modp
l),
if n E 0 (mod p  1).
roots mod p is congruent
9. If p is an odd prime > 3 prove that the product congruent to 1 mod p.
to p(p  1) mod p.
of the primitive
roots mod p
is
roots 10. Let p be an odd prime of the form 22k + 1. Prove that the set of primitive mod p is equal to the set of quadratic nonresidues mod p. Use this result to prove that 7 is a primitive root of every such prime.
222
Exercises for Chapter
11. Assume d (q(m). If d = exp,(a) we say that a is a primitive
10
root of the congruence
xd = 1 (mod m). Prove that if the congruence Pcrn) = 1 (mod m) has a primitive
root then it has &p(m))
12. Prove the properties
primitive
of indices described
13. Let p be an odd prime.
roots, incongruent
in Theorem
mod m.
10.10.
If (h, p) = 1 let
S(h) = {h”: 1 < n I cp(p  l), (n, p  1) = 1). If h is a primitive root of p the numbers in the set S(h) are distinct mod p (they are, in fact, the primitive roots of p). Prove that there is an integer h, not a primitive root of p, such that the numbers in S(h) are distinct mod p if, and only if, p = 3 (mod 4). 14. If m > 1 let pl, . . . . pk be the distinct prime divisors of cp(m). If (g, m) = 1 prove that g is a primitive root of m if, and only if, g does not satisfy any of the congruences gqp(m)ipg= 1 (mod m) for i = 1, 2, . . . , k. 15. The prime p = 71 has 7 as a primitive root. Find all primitive find a primitive root for p2 and for 2~‘. 16. Solve each of the following
roots of 71 and also
congruences:
(a) 8x = 7 (mod 43). (b) x8 = 17 (mod 43). (c) 8” = 3 (mod 43). 17. Let q be an odd prime
and suppose that p = 4q + 1 is also prime.
(a) Prove that the congruence x2 =  1 (mod p) has exactly two solutions, each of which is quadratic nonresidue of p. (b) Prove that every quadratic nonresidue of p is a primitive root of p, with the exception of the two nonresidues in (a). (c) Find all the primitive roots of 29. 18. (Extension of Exercise 17.) Let q be an odd prime and suppose that p = 2”q + 1 is prime. Prove that every quadratic nonresidue a of p is a primitive root of p if u*” f 1 (mod p). 19. Prove that there are only two real primitive showing their values. 24). Let x be a real primitive the form
character
characters
mod 8 and make
a table
mod m. If m is not a power of 2 prove that m has m = 2”p, . . . p,
where the pi are distinct
odd primes and CI = 0,2, or 3. If o! = 0 show that x(l)
and find a corresponding
formula
= ,(,,(pl)/* plm for x(  1) when CL= 2.
223
11
Dirichlet Series and Euler Products
11.1 Introduction In 1737 Euler proved Euclid’s theorem on the existence of infinitely many primes by showing that the series c p ‘, extended over all primes, diverges. He deduced this from the fact that the zeta function T(s),given by
for real s > 1, tends to cc as s * 1. In 1837 Dirichlet proved his celebrated theorem on primes in arithmetical progressions by studying the series (2)
L(s,x) = f 9 n=l
where x is a Dirichlet character and s > 1. The series in (1) and (2) are examples of series of the form (3) wheref(n) is an arithmetical function. These are called Dirichlet series with coefficients f(n). They constitute one of the most useful tools in analytic number theory. This chapter studies general properties of Dirichlet series. The next chapter makes a more detailed study of the Riemann zeta function C(s)and the Dirichlet Lfunctions L(s, x). 224
11.2:
Notation Following
The halfplane of absoluteconvergenceof a Dirichlet series
Riemann, we let s be a complex variable and write s = Q + it,
where Q and t are real. Then nS= es log” = e’“+“““~” = naeir log”. This shows that Ins1 = n” since leiel = 1 for real 8. The set of points s = cr + it such that o > a is called a halfplane. We will show that for each Dirichlet series there is a halfplane cr > gCin which the series converges, and another halfplane cr > a, in which it converges absolutely. We will also show that in the halfplane of convergence the series represents an analytic function of the complex variable s.
11.2 The halfplane of absolute convergence of a Dirichlet series First we note that if r~ 2 a we have InSI = nb 2 n” hence
f(n) < .If(
I if I  n” Therefore, if a Dirichlet series 1 f(n)n’ converges absolutely for s = a + ib, then by the comparison test it also converges absolutely for all s with cr 2 a. This observation implies the following theorem. Theorem 11.1 Suppose the series c (f(n)n“I does not converge for all s or diverge for all s. Then there exists a real number oa, culled the abscissa of absolute convergence, such that the series 1 f (n)n’ converges absolutely if o > (r,, but does not converge absolutely if cs < a,,.
PROOF.Let D be the set of all real (r such that c 1f (n)n‘1 diverges. D is not empty because the series does not converge for all s, and D is bounded above because the series does not diverge for all s. Therefore D has a least upper bound which we call cr,. If Q < a, then Q E D, otherwise crwould be an upper bound for D smaller than the least upper bound. If o > CJ.then D $ D since o, is an upper bound for D. This proves the theorem. 0 Note. If 1 I f(n)n“1 converges everywhere we define ca = co. If the series C If (n)n‘1 converges nowhere we define o’. = + co.
1 Riemann zeta function. The Dirichlet series I.“= 1 n’ converges absolutely for cr > 1. When s = 1 the series diverges, so CT,= 1. The sum of this series is denoted by c(s) and is called the Riemann zeta function. EXAMPLE
EXAMPLE 2 If f is bounded, say 1f(n) 1I A4 for all n 2 1, then c f (n)n’ converges absolutely for Q > 1, so 0. I 1. In particular if x is a Dirichlet character the Lseries L(s, x) = 2 x(n)n’ converges absolutely for g > 1.
225
11: Dirichlet series and Euler produets
EXAMPLE
3 The series c n”n’ diverges for every s so o. = + co.
EXAMPLE
4 The series 1 rr“n’ converges absolutely for every s so o, =  co.
11.3 The function series Assume that 1 f(n)n’ the sum function
defined by a Dirichlet converges absolutely for r~ > rro and let F(s) denote
(4) This section derives some properties of F(s). First we prove the following lemma. Lemma 1 Zf N 2 1 and 0 2 c > fra we have
PROOF.We have
The next theorem describes the behavior of F(s) as (T+ + co. Theorem
11.2 Zf F(s) is given by (4), then
lim F(a + it) = f(1) U+x, uniformly for  co < t < + ccl. Since F(s) = f(1) + cz=, f(n)nes we need only prove that the second term tends to 0 as 0 + + co. Choose c > cr,. Then for cr 2 c the lemma implies
PROOF.
where A is independent of a and t. Since A/2” + 0 as a + + 00 this proves 0 the theorem.
226
11.3 : The function
defined by a Dirichlet
series
We prove next that all the coefficients are uniquely determined by the sum function. Theorem 11.3 Uniqueness theorem. Given two Dirichlet series
F(s) = f f(n) n=l ns
andG(s)=
f
@,
n=l ns
both absolutely convergentfor o > oa. If F(s) = G(s)for each s in an infinite sequence {sk} such that ok + + co as k + co, then f(n) = g(n)for euery n. PROOF. Let h(n) = f(n)  g(n) and let H(s) = F(s)  G(s). Then H(sJ = 0
for each k. To prove that h(n) = 0 for all n we assume that h(n) # 0 for some n and obtain a contradiction. Let N be the smallest integer for which h(n) # 0. Then
.,s)=“~N!$L!g+
f h(n), n=~+l n
Hence h(N) = NW(s)  N” f
h(n).
n=N+l
ns
Putting s = sk we have H(sk) = 0 hence h(N) = N”’
‘f
h(n)n“k.
n=N+l
Choose k so that ok > c where c > o,. Then Lemma 1 implies Ih(
I N”*(N + l)(ukc) “E$+Ilh(n)lnc
=
where A is independent of k. Letting k + co we find (N/(N + 1))“L + 0 so h(N) = 0, a contradiction. cl The uniqueness theorem implies the existence of a halfplane in which a Dirichlet series does not vanish (unless, of course, the series vanishes identically). 11.4 Let F(s) = 1 f(n)n” and assume that F(s) # 0 for some s with a > aa. Then there is a halfplane o > c 2 o. in which F(s) is never zero.
Theorem
PROOF. Assume no such halfplane exists. Then for every k = 1, 2, . . . there
is a point sk with crk > k such that F(sJ = 0. Since (TV+ + cc as k + 00 the uniqueness theorem shows thatf (n) = 0 for all n, contradicting the hypothesis that F(s) # 0 for some s. cl 227
11: Dirichlet
series and Euler products
11.4 Multiplication
of Dirichlet series
The next theorem relates products of Dirichlet series with the Dirichlet convolution of their coefficients. Theorem 11.5 Given twofunctions
F(s) and G(s) represented by Dirichlet
F(s) = fj fO n=l nS
for 0 > a>
m s(n) G(s) = 1 s n=l n
for o > b.
series,
and
Then in the halfplane
where both series converge absolutely we have F(s)G(s) = f h(n), n=l ns
(5)
where h = f * g, the Dirichlet
convolution
Conversely, ifF(s)G(s) = c a(n)n‘for as k + co then a = f * g. PROOF.
off and g:
all s in a sequence {sk} with ok + + co
For any s for which both series converge absolutely we have FW(s)
= $If
(n)nSm$Ig(m)mS
= f
f
f (n)g(m)(mn)‘.
n=lm=l
Because of absolute convergence we can multiply these series together and rearrange the terms in any way we please without altering the sum. Collect together those terms for which mn is constant, say mn = k. The possible values ofkare1,2,...,hence WW
= $~~J+M4)’
= $~(k)k”
where h(k) = xntnck f (n)g(m) = (f * g)(k). This proves the first assertion, and the second follows from the uniqueness theorem. Cl EXAMPLE 1 Both series 1 n’ and c p(n)n” converge absolutely for 0 > 1. Taking f (n) = 1 and g(n) = p(n) in (5) we find h(n) = [l/n], so
((s)f
n=l
228
!!f@= 1 ifo> nS
1.
11.4: Multiplication
of Dirichlet
series
In particular, this shows that i(s) # 0 for cr > 1 and that
EXAMPLE 2 More generally, assume f( 1) # 0 and let g = fi, the Dirichlet inverse off. Then in any halfplane where both series F(s) = 1 ,f(n)n” and G(s) = c g(n)n ’ converge absolutely we have F(s) # 0 and G(s) = l/F(s). EXAMPLE 3 Assume F(s) = 1 f(n)n” converges absolutely for d > go. If f is completely multiplicative we havef  ‘(n) = &)f(n). Since 1f  l(n)1 I a1so converges absolutely for 0 > eraand we 1f(n) 1the series 1 &)f(n)n” have 1O” PM(4 ;I;=& ifo>a,. n=l
In particular for every Dirichlet character x we have
mAMn) 1 cyp=Lb,xl
ifa > 1.
n=l
EXAMPLE 4 Take f(n) = 1 and g(n) = q(n), Euler’s totient. Since q(n) < n converges absolutely for a > 2. Also, h(n) = Lln q(d) the series 1 cp(n)nWS = n so (5) gives us Qs) f co(n) = “zl a = [(s  1) if a > 2. n=l ns Therefore jr?=?
ifa>2.
EXAJWLE 5 Take f(n) = 1 and g(n) = n’. Then h(n) = zln d” = a,(n), and (5) gives us [(s)[(s  a) = f
y
if a > max{l, 1 + Re(a)}.
n=l
EXAMPLE 6 Takef(n)
= 1 and g(n) = A(n), Liouville’s function. Then
1 if n = m* for some m, h(n) = 1 A(d) = 0 otherwise 3 din so (5) gives us
229
11: Dirichlet
series and Euler products
Hence
11.5 Euler products The next theorem, discovered by Euler in 1737, is sometimes called the analytic version of the fundamental theorem of arithmetic. Theorem 11.6 Let f be a multiplicative
arithmeticalfunction such that the series 1 f(n) is absolutely convergent. Then the sum of the series can be expressed as an absolutely convergent infinite product, (6)
“Elf (4 =
nu +f (PI + f V)
+ . ..I
11
extended over all primes. If f is completely simplifies and we have
multiplicative,
the product
Note. In each case the product is called the Euler product of the series. PROOF.
Consider the finite product P(x) = nil PSX
+ f(P) + f (P’) + . . .>
extended over all primes p I x. Since this is the product of a finite number of absolutely convergent series we can multiply the series and rearrange the terms in any fashion without altering the sum. A typical term is of the form f (Pl”‘)f
since f is multiplicative. write
(P2”?
. . . f (P,“‘) = f
f.P1a’P2a*
. . . P?)
By the fundamental theorem of arithmetic we can fW
= 1 f(n) IlEA
where A consists of those n having all their prime factors 5 x. Therefore n$If (4  p(x) = nTBf (n),
where B is the set of n having at least one prime factor >x. Therefore
As x + co the last sum on the right ) 0 since 11 f(n) 1is convergent. Hence P(x) + 1 f(n) as x + co.
230
11.5:
Euler
products
Now an infinite product of the form n (1 + a,) converges absolutely whenever the corresponding series 1 a, converges absolutely. In this case we have p;xl fw + a21 + . . .I
on. Iff
is
we have
It should be noted that the general term of the product in (8) is the Bell seriesf,(x) of the functionfwith x = p‘. (See Section 2.16.) Takingf(n) = 1, p(n), q(n), a,(n), L(n) and x(n), respectively, we obtain the following Euler products :
EXAMPLES
231
11: Dirichlet
series and Euler products
Note. If x = xi, the principal character mod k, then xi(p) = 0 if plk and xi(p) = 1 if p ,l’ k, so the Euler product for L(s, xi) becomes
L(s,Xl)= n ___1 =
n (1  p“) = i(s) n (1  p+). plk Plk Thus the Lfunction L(s, xi) is equal to the zeta function i(s) multiplied by a finite number of factors. PXk
1

P”
____. l
II p
1 
P’
11.6 The halfplane of convergence of a Dirichlet series To prove the existence of a halfplane of convergence we use the following lemma. Lemma 2 Let so = (r. + it, and assume that the Dirichlet has bounded partial sums, say IEfbb“‘~
series 1 f(n)nso
5 M
for all x 2 1. Then for each s with CJ> o. we have
a 0. In particular, this shows that the Dirichlet series f t11 “=I ns converges for cr > 0 since 1En,, (  1)” 15 1. Similarly, if 1 is any nonprincipal Dirichlet character mod k we have 1xnI x x(n) ) 5 q(k) so
converges for r~ > 0. The same type of reasoning gives the following theorem. Theorem 11.8 Zf the series 1 f(n)n”
converges for s = CT~+ it, then it also converges for all s with Q > oO. If it diverges for s = crO + it, then it diverges for all s with o < CT,,.
PROOF. The second statement follows from the first. To prove the first statement, choose any s with cr > eO. Lemma 2 shows that
where K is independent of a. Since aaoMa ) 0 as a P + 03, the Cauchy condition shows that 1 f (n)n’ converges. cl Theorem 11.9 If the series c f(n)n’
d oes not converge everywhere or diverge everywhere, then there exists a real number oC, called the abscissa of convergence, such that the series converges for all s in the halfplane o > ue and diverges for all s in the halfplane CT< cc.
PROOF. We argue as in the proof of Theorem 11.1, taking cr, to be the least upper bound of all Q for which c f(n)n’ diverges. 0
Note. If the series converges everywhere we define ec = co, and if it converges nowhere we define tsr = + 00.
Since absolute converge implies convergence, we always have era2 ec. If 6, > cr, there is an infinite strip cr, < r~ < o, in which the series converges conditionally (see Figure 11.1.) The next theorem shows that the width of this strip does not exceed 1.
0,
Figure 11.1 233
11: Dirichlet
series and Euler products
Theorem 11.10 For
any Dirichlet
series with o,jnite
we have
PROOF. It suffices to show that if 1 f(n)n“O converges for some s0 then it converges absolutely for all s with (r > o0 + 1. Let A be an upper bound for the numbers )f(n)n“Oj . Then
so c 1f(n)n“1
converges by comparison with c naoa.
Cl
EXAMPLE The series
f n=l
(1) ns
converges if o > 0, but the convergence is absolute only if (r > 1. Therefore in this example cc = 0 and ca =: 1. Convergence properties of Dirichlet series can be compared with those of power series. Every power series has a disk of convergence, whereas every Dirichlet series has a halfplane of convergence. For power series the interior of the disk of convergence is also the domain of absolute convergence. For Dirichlet series the domain of absolute convergence may be a proper subset of the domain of convergence. A power series represents an analytic function inside its disk of convergence. We show next that a Dirichlet series represents an analytic function inside its halfplane of convergence.
11.7 Analytic properties of Dirichlet series Analytic properties of Dirichlet series will be deduced from the following general theorem of complex function theory which we state as a lemma. Lemma 3 Let { fn} be a sequence of functions
analytic on an open subset S of the complex plane, and assume that { fn} converges uniformly on every compact subset of S to a limit function J Then f is analytic on S and the sequence of derivatives {f b} converges uniformly on every compact subset of S to the derivative f I.
PROOF.
234
Sincefn is analytic on S we have Cauchy’s integral formula
11.7 : Analytic
properties
of Dirichlet
series
where D is any compact disk in S, c?D is its positively oriented boundary, and a is any interior point of D. Because of uniform convergence we can pass to the limit under the integral sign and obtain
which implies thatf is analytic inside D. For the derivatives we have
from which it follows easily that x(a) + j‘(a) uniformly on every compact subsetofSasn+ cc. 0 To apply the lemma to Dirichlet series we show first that we have uniform convergence on compact subsets of the halfplane of convergence. Theorem 11.11 A Dirichlet series 1 f(n)Ks converges uniformly on every compact subset lying interior to the halfplane of convergence o > gC.
PROOF.It suffices to show that 1 f(n)Cs converges uniformly on every compact rectangle R = [a, fl] x [c, d] with a > cc. To do this we use the estimate obtained in Lemma 2,
where s,, = o,, + it, is any point in the halfplane 0 > oc and s is any point with 0 > oO. We choose s0 = (TV where oc < o0 < a. (See Figure 11.2.)
Then if s E R we have 0  o,, 2 a  c0 and 1s0  s 1< C, where C is a constant depending on s0 and R but not on s. Then (10) implies
a o. we have
In particular,
(a) lim 
1
~rmzT
converges
absolutely
for
c > C.
if o > 1 we have
T
I &a + it) 1’ dt = “El & = ((20).
s T
~+rnzT
(b) lim 
f(n)n”
1
T
Ic(")(a
+
it)12
s T
(c) li+i&Jy
dt
=
f
k.$f
=
Q92o).
n=l
Il(a+it)lm2dt= T
f
$$=g.
a
n=l
I&a+it)14dt=
f F=$$$. II=1
Formula (18) follows by taking g(n) = f(n) in Theorem 11.15. To deduce the special formulas (a) through(d) we need only evaluate the Dirichlet series 1 If(n)IZn2u for the following choices off(n): (a) f(n) = 1; (b) f(n) = ( l)k logk n; (c) f(n) = p(n); (d) f(n) = a,(n). The formula (a) is clear, and formula (b) follows from the relation PROOF.
c’k+) = ( l)k f
logk
.
n=l nS To prove (c) and (d) we use Euler products. For (c) we 6ave
m P2(n) n (1 + ps) = n 1  Pe2”_ GS) c=,ns p 1 p” n=l aw. Replacing s by 2a we get (c). For (d) we write = n { 1 + ao2(P)p” + a02(p2)p2S + . . .) cO”ao2(n) ns
n=l
P
= rj (1 + 22p” + 32p2” + . . .} / J =n{~(n+l)2~“sj=~(:I:~~~4=~ n=O P
\oP
1  x2 x+1 since C.“=o (n + 1)2x” = ~ =~ . Now replace s by 2a to get (d). (x  1)3 (1 x)4 0 241
11: Dirichlet
series and Euler products
11.11 An integral formula for the coefficients of a Dirichlet series Theorem 11.17 Assume the series F(s) = ~.“=lf(n)n” for a > a,. Then for a > a0 and x > 0 we have F(a t it)xO+” dt =
converges absolutely
f(n) o
ifx = n, otherwise,
PROOF. For a > a0 we have
= eit s
log(x/n)
&,
T
since the series is uniformly convergent for all t in any interval [  7; T]. If x is not an integer then x/n # 1 for all n and we have
s T
2 sin
,it lo&/n) dt =
T
and the series becomes
[ 01
sin T log 5 n
which tends to 0 as T + co. However, if x is an integer, say x = k, then the term in (19) with n = k contributes
and hence
The second term tends to 0 as T + co as was shown in first part of the 0 argument.
11.12 : An integral
formula
for the partial
sums of a Dirichlet
series
11.12 An integral formula for the partial sums of a Dirichlet series In this section we derive a formula of Perron for expressing the partial sums of a Dirichlet series as an integral of the sum function. We shall require a lemma on contour integrals. Lemma 4 Zfc > 0, define j:‘zi real number, we have
to mean lim,,,
Js?iF. Then ifa is any positive
1 ifa> 1
ifa = 1,
Z
if0 < a < 1.
0 Moreover,
1,
we have
~,:i&‘“~~nTl~gcLi
ifO 1 we use instead the contour R shown in Figure 11.5. Here b > c > 0 and T > c. Now a’/z has a first order pole at z = 0 with residue 1 since az = ez loga = 1 + zloga + O(IzI’) asz+O.
Therefore
hence
We now estimate the integrals on the right. We have
Is,l:ri~az~~l~~~b~~~~.
As b + co the second integral tends to 0 and we obtain (21). 244
11.12: An integral formula for the partial sumsof a Dirichlet series When a = 1, we can treat the integral directly. We have
= 2ic
T dy I JTy+
the other integral vanishing because the integrand is an odd function. Hence T 1 1  arctan  =    arctan 2 c 2 71 T’ Since arctan c/T < c/T this proves (22), and the proof of Lemma 4 is complete. 0 11.18 Perron’s formula. Let F(s) = c.“=I f(n)/n’ be absolutely convergentfor CJ> a,; let c > 0, x > 0 be arbitrary. Then ifa > aa  c we have :
Theorem
where I* means that the last term in the sum must be multiplied by l/2 when x is an integer. PROOF.In the integral, c is the real part of z, so the series for F(s + z) is absolutely and uniformly convergent on compact subsets of the halfplane a + c > a,. Therefore
s c+iT
ciT
, , f(x) XS
c+iT dz 3 sc,iT Z
the symbol +’ indicating that the last term appears only if x is an integer. In the finite sum En cx we can pass to the limit T + cc term by term, and the integral is 27ciby Lemma 4. (Here a = x/n, a > 1.)The last term (if it appears) yields @(x)x’ and the theorem will be proved if we show that
245
11 : Dirichlet
series and Euler products
We know that i:?iz (x/n)z(dz/z) = 0 if n > x but to prove (23) we must estimate the rate at which is’$ tends to zero. From Lemma 4 we have the estimate if0 < a < 1.
Here a = x/n, with n > x. In fact, n 2 [x] + 1, so that l/a = n/x 2 ([x] + 1)/x. Hence + 1) X
asT+co.
0
This proves Perron’s formula.
Note. If c > (T, Perron’s formula is valid for s = 0 and we obtain the following integral representation for the partial sums of the coefficients:
Exercises for Chapter 1. Derive the following (a) c(s) = s lIK $
11
identities, dx.
(b); f =sjlax%dx,
where the sum is extended
dx,
(d)  ;
= s j,% !$
(e) L(s, x) = s jlm $
valid for 0 > 1.
where M(x) = c p(n). nsx dx,
where $(x) = 1 A(n). n5.x
dx,
where A(x) = 1 x(n). “5.X
Show that (e) is also valid Theorem 4.2.1 246
over all primes.
for 0 > 0 if x is a nonprincipal
character.
[Hint:
Exercises for Chapter
2. Assume that the series x2=1 f( n ) converges with sum A, and let A(x) = In,, (a) Prove that the Dirichlet 0 > 0 and that
series F(s) = c.“=,
f(n)n’
11
f(n).
converges for each s with
where R(x) = A  A(x). [Hint: Theorem 4.2.1 (b) Deduce that F(o) + A as (r + O+. (c) If 0 > 0 and N 2 1 is an integer, prove that
m 4~) s NY s+l dY. (d) Write
s = 0 + it, take N = 1 + [ 1t I] in part (c) and show that IF(u + it)1 = O(ltl’“)
if0 < 0 < 1.
3. (a) Prove that the series 1 n 1 if has bounded partial sums if t # 0. When t = 0 the partial sums are unbounded. (b) Prove that the series 1 n ‘“diverges for all real t. In other words, the Dirichlet series for c(s) diverges everywhere on the line u = 1. 4. Let F(s) = CT=, ,f(n)nms where ,f(n) is completely multiplicative converges absolutely for 0 > co. Prove that if u > go we have F’(s) =F(s)
and the series
f f(n)W 7’ “=,
In the following exercises, A(n) is Liouville’s function, d(n) is the number of divisors of n, v(n) and I are defined as follows: v(1) = 0, ~(1) = 1; if n = pin’ . . . pkakthen v(n) = k and x(n) = ~2~~1~ . . . ak. Prove that the identities in Exercises 5 through 10 are valid for 0 > 1.
, f 2”” = 1*(s) n=l ns a24 11. Express the sum of the series I.“=, function. 12. Letfbe a completely If the series 1 f(n)nes F(s) # 0 and that
3v%(n)~(n)ns
in terms of the Riemann
zeta
multiplicative function such thatf(p) = f(p)* for each prime p. converges absolutely for 0 > Q, and has sum F(s), prove that
f fol(n) “=I
ns
= FW 
if0>0,.
F(s)
247
11: Dirichlet series and Euler products
13. LetSbe a multiplicative function. such thatf(p)
c &t)f(n)n” and that
= f(p)* for each prime p. If the series converges absolutely for cr > a, and has sum F(s), prove that F(s) # 0
14. Let f be a multiplicative
function such that 1 f(n)nms converges absolutely for Q > 0,. If p is prime and cr > ua prove that
where p(p, n) is the Mobius function evaluated at the gcd of p and n. [Hint: Euler products.] 15. Prove that
More generally, if each si has real part bi > 1, express the multiple sum ~~,...nE,ml’~...m.l, r onI..... In,,=1
in terms of the Riemann zeta function. 16. Integrals of the form f(s) = jlm F
(24)
dx,
where A(x) is Riemannintegra.ble on every compact interval [l, a], have some properties analogous to those of Dirichlet series. For example, they possess a halfplane of absolute convergence u > Q, and a halfplane of convergence Q > cC in which ,f(s) is analytic. This exercise describes an analogue of Theorem 11.13 (Landau’s
theorem).
Let f’(s) be represented in the halfplane rr > Q, by (24), where a, is finite, and assume that A(x) is realvalued and does not change sign for x 2 x0. Prove that f(s) has a singularity on the real axis at the point s = u,. 17. Let l,(n) = Lln d”l(d) where A(n) is Liouville’s max{l, Re(rr) + 1}, we have “E, “$4 _ I(sK&w and
248
function. Prove that if Q >
The Functions 5(s) and WY x)
12
12.1 Introduction This chapter develops further properties of the Riemann zeta function c(s) and the Dirichlet Lfunctions L(s, x) defined for cr > 1 by the series and&X)=
f @. n=l ns
As in the last chapter we write s = r~ + it. The treatment of both C(s)and L(s, x) can be unified by introducing the Hurwitz zeta function ((s, a), defined for cr > 1 by the series
as,4 = “ZO&. Here a is a fixed real number, 0 < a I 1. When a = 1 this reduces to the Riemann zeta function, C(s)= c(s, 1). We can also express L(s, x) in terms of Hurwitz zeta functions. If x is a character mod k we rearrange the terms in the series for L(s, x) according to the residue classes mod k. That is, we write n=qk+r, where1 0 can be continued beyond the line (i = 0, and F(s) exists as a function which is analytic everywhere in the splane except for simple poles at the points s = o:,  1,  2,  3, . . . ) with residue ( l)“/n ! at s = rt. We also have the representation nsn !
r(s) = “+mS(S+ lim ~ l)...(s+n)
fors+Q1,2,...,
and the product formula 1 r(s)

SeCs (D
1 + f
“:E 41
esl”
>
for all s,
where C is Euler’s constant. Since the product converges for all s, T’(s) is never zero. The gamma function satisfies two functional equations, (2)
rjs + 1) = dts)
and
r(s)r(i  S) = &, valid for all s, and a multiplication (4)
formula
r~sjr(s + Jf... r(s + G)
valid for all s and all integers m 2 1. 250
= (2n)(m1)IZm(1/2)msr(mS),
12.3: Integral representation for the Hurwitz zeta function
We will use the integral representation (l), the functional equations (2) and (3), and the fact that T(s) exists in the whole plane, with simple poles at the integers s = 0,  1, 2, . . . We also note that I+ + 1) = n ! if n is a nonnegative integer.
12.3 Integral representation for the Hurwitz zeta function The Hurwitz
zeta function
[(s, a) is initially cc
defined for 0 > 1 by the series 1
Theorem 12.1 The series for [(s, a) conuerges absolutely for CY> 1. The convergence is uniform in every halfplane CT2 1 + 6, 6 > 0, so [(s, a) is an analytic function of s in the halfplane (T > 1. PROOF.
All these statements follow from the inequalities g*
Theorem
n + a)‘[
= 1 (n + ~2~~ I n=1
1 (n + a)(‘+@.
0
fl=l
12.2 For o > 1 we have the integral representation
In particular,
when a = 1 we have I(s)((s) = Jam G
dx.
PROOF. First we keep s real, s > 1, and then extend the result to complex s by analytic continuation. In the integral for I(s) we make the change of variable x = (n + a)t, where n 2 0, to obtain
r(s) =
emxxsml dx = (n + a)
(n + a)“r(s) Summing
=
s0
me“‘e“‘tSl
,@+a+
1 dt,
dt.
over all n 2 0 we find c(s, a)r(s) = f ~me“‘e‘Yn=O 0
’ dt, 251
12: The functions i(s) and L(s, x)
the series on the right being convergent if 0 > 1. Now we wish to interchange the sum and integral. The simplest way to justify this is to regard the integral as a Lebesgue integral. Since the integrand is nonnegative, Levi’s convergence theorem (Theorem 10.25 in Reference [2]) tells us that the series ,I) jy e“’ e la t sl “YO
converges almost everywhere to a sum function which is Lebesgueintegrable on [O, + co) and that ((s, a)r(s) = f /me~“‘e~“t’l n=O 0
dt = lorn n~oe“‘eartsml
dt.
Butift>OwehaveO<e‘ 1. To extend it to all complex s with CJ> 1 we note that both members are analytic for c > 1. To show that the right member is analytic we assume 1 + 6 I CT5 c, where c > 1 and 6 > 0 and write lom IGidt
5 I,“;sdt
= (IO1 + J;“)gdt.
If 0 I t I 1 we have to’ I t’, and if t 2 1 we have to l I tC l. Also, since e’  I 2 t for t 2 0 we have ‘La1 dt  ‘ia,
s0 and
st
m e“’ CT1
1
1

e’
at
’
This shows that the integral in (5) converges uniformly in every strip 1 + 6 5 c I c, where 6 > 0, and therefore represents an analytic function in every such strip, hence also in the halfplane 0 > 1. Therefore, by analytic continuation, (5) holds for all s with CJ> 1. Cl 252
12.4: A contour
integral
representation
for the Hunvitz
zeta function
12.4 A contour integral representation for the Hurwitz zeta function To extend [(s, a) beyond the line 0 = 1 we derive another representation in terms of a contour integral. The contour C is a loop around the negative real axis, as shown in Figure 12.1. The loop is composed of three parts C1, CZ, C3. C2 is a positively oriented circle of radius c < 2~ about the origin, and C,, C3 are the lower and upper edges of a “cut” in the zplane along the negative real axis, traversed as shown in Figure 12.1.
Figure
12.1
This means that we use the parametrizations z = re“’ on C1 and z = I@ on C3 where Yvaries from c to + co. Theorem 12.3 Zf 0 < a < 1 the function
dejined by the contour integral
is an entire function
of s. Moreover,
we have
(6)
((s, a) = r( 1  s)Z(s,a) if 0 > 1.
PROOF.Here zsmeans r’e“” on C1 and Yenis on C3. We consider an arbitrary compact disk 1s 1I M and prove that the integrals along C1 and C3 converge uniformly on every such disk. Since the integrand is an entire function of s this will prove that Z(s,a) is entire. Along C, we have, for r 2 1, Izsll
=
f11,ni(al+it)l
=
flent
I
rMlenM
since IsI < M. Similarly, along C3 we have, for r 2 1, IzslI
=
r~lleffi(ul+it)I
=
f7lent
I
@flenM.
Hence on either C, or C3 we have, for r 2 1, zsl euz rM lenMeor rMlenMe(l a)r I 1 e’ = e’1 ’ I 1  eZ I But e’  1 > e’/2 when r > log 2 so the integrand is bounded by A?’ epur where A is a constant depending on M but not on r. Since sf” rMpleCar dr converges if c > 0 this shows that the integrals along Cl and C3 converge 253
12 : The functions
c(s) and L(s, x)
uniformly on every compact disk 1st I M, and hence Z(s, a) is an entire function of s. To prove (6) we write
2niz(s,u) = (J,,+ J‘,,+ JcJz’%(r)dz where g(z) = eaz/(1  e’). On Ci and C, we have g(z) = g( r), and on C2 we write z = ceie, where 7~ < ~9I IL This gives us 2~il(~, u) =
d(j ’ , lenisg(  r) dr + i z cs le(s l)ieceieg(ceie) Im I 77 a, rs ‘enisg(  r) dr + ic m
= 2i sin(7rs)
n
rSl
sc
g(  r) dr + ic”
s II
eiseg(ceie) dB.
Dividing by 2i, we get nZ(s, a) = sin(rrs)Z,(s, c) + Z,(s, c) say. Now let c + 0. We find .‘m
lim Z,(s, c) = C+O
J
rsl
e
01
o 1  e’
dr = T(s)@, a),
if CJ> 1. We show next that lim,,, Z,(s, c) = 0. To do this note that g(z) is analytic in I z ( < 27r except for a first order pole at z = 0. Therefore zg(z) is analytic everywhere inside lz 1< 271 and hence is bounded there, say lg(z) 1i A/I z 1,where )z 1= c < 27~and A is a constant. Therefore we have (Z,(s, c)l I 5 /I,epre
$ d0 I Ae’%
‘.
If 0 > 1 and c + 0 we find Z,(s, c) + 0 hence rcZ(s,a) = sin(rcs)IJs)&, a). Since I(s)I(l  s) = rc/sin rcsthis proves (6). 0
12.5 The analytic continuation Hurwitz zeta function
of the
In the equation 5(s, a) = I(1  s)Z(s,a), valid for 0 > 1, the functions Z(s,a) and I( 1  s) are meaningful for every complex s. Therefore we can use this equation to define [(s, a) for r~ I 1. Definition (7)
If (T I 1 we define {(s, a) by the equation
((s,U)= r(i  S)Z(S, u).
This equation provides the analytic continuation splane. 254
of i(s, a) in the entire
12.6: Analytic continuation of c(s)and L(s, x) Theorem 12.4 The function Qs, a) so de$ned is analytic for all s except for a simple pole at s = 1 with residue 1. hmOF. Since I(s, a) is entire the only possible singularities of i(s, a) are the poles of I(1  s), that is, the points s = 1,2, 3, . . . But Theorem 12.1 shows that c(s, a) is analytic at s = 2, 3, . . . , so s = 1 is the only possible pole of 5th a). Now we show that there is a pole at s = 1 with residue 1. Ifs is any integer, say s = n, the integrand in the contour integral for Z(s, a) takes the same values on C, as on C3 and hence the integrals along C, and C3 cancel, leaving Zn lea2 yl1 IIZ ____ dz = Res e Z(n, 4 = & z=. 1  ez . I c2 1  ez
In particular when s = 1 we have 1(1,a) = Rese”‘= z=o 1  ez
lim zeDZ = lim f. = lim I =  1, r+O 1  e’ z+o 1  e’ 2O ez
To find the residue of ((s, a) at s = 1 we compute the limit lim(s  l)[(s, a) =  lim(1  s)I(l  s)l(s, a) =  Z(1, a)lim I(2  s) s+l s1 S+1 = r(l) = 1. This proves that Qs, a) has a simple pole at s = 1 with residue 1.
0
Note. Since c(s, a) is analytic at s = 2,3, . . . and I( 1  s) has poles at these points, Equation (7) implies that I(s, a) vanishes at these points.
12.6 Analytic continuation of c(s) and IQ, x) In the introduction
we proved that for o > 1 we have r(s) = as, 1)
and
L(s,x) =k“&r)is,;) r=1
(
)
where x is any Dirichlet character mod k. Now we use these formulas as de$nitions of the functions c(s)and L(s, x) for o I 1. In this way we obtain the analytic continuation of l(s) and L(s, x) beyond the line CT= 1. Theorem 12.5 (a) The Riemann zeta function c(s) is analytic everywhere exceptfor a simple pole at s = 1 with residue 1. (b) For the principal character x1 mod k, the Lfunction L(s, xl) is analytic everywhere except for a simple pole at s = 1 with residue cp(k)/k. (c) Zf x # xl, L(s, x) is an entire function
of s.
255
12: The functions C(S)and L(s, x)
PROOF. Part (a) follows at once from Theorem 12.4. To prove (b) and (c) we use the relation
Since I$, r/k) has a simple pole at s = 1with residue 1,the function X(r)&, r/k) has a simple pole at s = 1 with residue X(r). Therefore Res L(s, x) = lim(s  l)L(s, x) = lim(s  l)k” f: x(r)l s=l
S+1
Sl
&
r=l
ifx
#
x1,
if*
=
x1
0
k
12.7 Hurwitz’s formula for [(s, a) The function [(s, a) was originally defined for cr > 1 by an infinite series. Hurwitz obtained another series representation for [(s, a) valid in the halfplane u < 0. Before we state this formula we discuss a lemma that will be used in its proof. Lemma 1 Let S(r) denote the region that remains when we remove from the zplane all open circular disks of radius r, 0 < r < 71, with centers at z = 2imi,n = 0, +l, +2 ,... Then if0 < a I 1 thefunction
is bounded in S(r). (The bound depends on r.) PROOF. Write z = x + iy and consider the punctured rectangle Q(r) = {z: 1x1 I
1, lyl I 71,IzI 2 r},
shown in Figure 12.2.
Figure 12.2 256
12.7 : Hurwitz’s formula for [(s, a)
This is a compact set so g is bounded on Q(v). Also, since 1g(z + 274 I = 1g(z) 1,g is bounded in the punctured infinite strip {z:lxl
I 1,lz  2nnil 2 r,n = 0, fl,
+2 )... }.
Now we show that g is bounded outside this strip. Suppose 1x1 2 1 and consider
Forx>lwehave\le”l=eXlande”“<e”,so
Also,whenx 1. Note that F(x, s) is a periodic function of x with period 1 and that F(1, s) = c(s). The series converges absolutely if Q > 1. If x is not an integer the series also converges (conditionally) for 0 > 0 because for each fixed nonintegral x the coefficients have bounded partial sums. Note. We shall refer to F(x, s) as the periodic zetafunction. Theorem
12.6 Hurwitz’s formula. ZJO < a I 1 and IT > 1 we have
Us) {e nrs’2Z$q Cl1 s,4 = (271)” Zf a # 1 this representation PROOF.
s) + eXiS12F( a, s)}.
is also valid for 0 > 0.
Consider the function zsl oz zhb, 4 = &
where C(N) is the contour
t
I C(N) 1  ez
shown in Figure
dz,
12.3, N being an integer. 257
12 : The functions
c(s) and L(s, x)
l
(2N + 2)ni
1R=(2N+l)r
Figure
12.3
First we prove that lim,,, I N(s, a) = Z(s,a) if (T < 0. For this it suffices to show that the integral along the outer circle tends to 0 as N + co. On the outer circle we have z = Reie, 71 I 8 I 71,hence lzS
1) =
IRS
leiL%
1)I
=
R”
lete
and this + 0 as R + co if 0 < 0. Therefore, replacing s by 1  s we see that lim Z,(l  s, a) = I(1  s, a) if g > 1.
(11)
NCC
Now we compute I,( 1  s, a) explicitly by Cauchy’s residue theorem. We have I,(1  s, a) = 
; R(n) =  f {R(n) + R( n)> “cN n=l n#O
where
Now e2nnia
R(n) = lim (z  2nni) 2; z
258
2nni
= (2mi) ,yEzi
e2nnio z  2mi __ 1  ez = (2nni)” ’
12.8 : The functional equation for the Riemann zeta function
hence N INtl

s,
‘)
=
al
=
,+nio
n;,
h/2 zN(l

‘2
N
;2,q
e
@&$
“z,
+
e2nnia
Znnia
f
n=l (2n7ci)“’
pis/Z
N
ns + (27~)“~~~ 1
e
Znnia
nS ’
Letting N + co and using (11) we obtain  nis/2
nis/Z
I( 1  s, a) = e(2n)”
F(a, s) + k
F(  a, s).
Hence
Us) {e  nrs’2F(u, s) + enis”F(4 [(I  S, U) = T(~)z(l  S, a) = o”
12.8 The functional equation Riemann zeta function The first application for i(s).
of Hurwitz’s
s)).
0
for the formula
is Riemann’s
functional
equation
Theorem 12.7 For all s we have
or, equivalently, c(s) = 2(2n)“ ‘I71  s)sin 7 0
(13) PROOF. Taking
a = 1 in the Hurwitz
formula
r(s) {e  n’“‘2[(s) + e”““i(s)}
[(l  s) = oS
[(l  s).
we obtain, for (T > 1,
r(s) 2 cos
= o”
y
 c(s).
0 This proves (12) for o > 1 and the result holds for all s by analytic continuation. To deduce (13) from (12) replace s by 1  s. 0
Note. Taking s = 2n + 1 in (12) where n = 1,2,3, . . . , the factor cos(ns/2) vanishes and we find the socalled trivial zeros of C(s), c(2n)=O
fern=
1,2,3,... 259
12 :
The functions c(s)and L(s, x)
The functional equation can be put in a simpler form if we use Legendre’s duplication formula for the gamma function, 27+2T(2s)
= T(s)I
s+ ; (
, >
which is the special case m = 2 of Equation (4). When s is replaced by (1  s)/2 this becomes
Since
this gives us
qi s)siny=
2snl/2r ls ( 2 ). ri 0
Using this to replace the product I( 1  s)sin(rrs/2) in (13) we obtain
ns/2ri [cs)= 71(1Wrq Ql ( > 0
 s).
In other words, the functional equation takes the form o(s) = o(l  s), where
a+) = 71s12ri C(S). 0 The function Q(s) has simple poles at s = 0 and s = 1. Following Riemann, we multiply Q(s) by s(s  1)/2 to remove the poles and define 5(s) = ; s(s  l)@(s). Then l(s) is an entire function of s and satisfies the functional equation 5(s) = 5(1  s). 260
12.10: The functional
equation
for Lfunctions
12.9 A functional equation for the Hurwitz zeta function The functional equation for c(s) is a special case of a functional equation for &s, a) when a is rational. Theorem 12.8 Zf h and k are integers, 1 I h I k, then for all s we have (14)
r s, f . >( >
r
PROOF. This comes from the fact that the function F(x, s) is a linear combination of Hurwitz zeta functions when x is rational. In fact, if x = h/k we can rearrange the terms in (9) according to the residue classesmod k by writing
n = qk + r,
where 1 I r I k and q = 0, 1,2, . . .
This gives us, for o > 1,
.
Therefore if we take a = h/k in Hurwitz’s formula we obtain
=
which proves (14) for o > 1.The result holds for all s by analytic continuation. cl It should be noted that when h = k = 1 there is only one term in the sum in (14) and we obtain Riemann’s functional equation.
12.10 The functional equation for Lfunctions Hurwitz’s formula can also be used to deduce a functional equation for the Dirichlet Lfunctions. First we show that it suffices to consider only the primitive characters mod k. 261
12 : The functions
i(s) and L(s, x)
Theorem 12.9 Let x be any Dirichlet modulus, and write
character
mod k, let d be any induced
x(n) = $(4x,(n), where Ic/ is a character mod d and x1 is the Principal character mod k. Then for all s we have
L(s, x) = L(s, Ic/)n 1 elk PROOF.
First keep o > 1 and use the Euler product L(s, x) = Jj ~. “lx@)
1 P”
Since x(p) = t+Q)xi(p) and since xi(p) = 0 if p 1k and xi(p) = 1 if p ,j’ k we find
This proves the theorem for 0 > 1 and we extend it to all s by analytic 0 continuation. Note. If we choose d in the foregoing theorem to be the conductor of x, then $ is a primitive character modulo d. This shows that every Lseries L(s, x) is equal to the Lseries L(s, @) of a primitive character, multiplied by a finite number of factors. To deduce the functional equation for Lfunctions from Hurwitz’s formula we first express L(s, x) in terms of the periodic zeta function F(x, s). Theorem 12.10 Let x be a primitive
character mod k. Thenfor
GU, X)W, x) = i SW h=l
where G(m, x) is the Gauss sum associated with x, G(m, x) = i X(r)eZnirmlk. r=l
262
,
rs > 1 we have
12.10: The functional
PROOF.
equation
for Lfunctions
Take x = h/k in (9), multiply by j(h) and sum on h to obtain
= c nG(n, j). n=l
But G(n, j) is separable because j is primitive, so G(n, 2) = &)G(l,
j)), hence 0
GU, Xl f x(n)n” = ‘31, XWb, xl. n=l
Theorem 12.11 Functional equation for Dirichlet primitive character mod k then for all s we have
(16)
k” ‘I(s)
L(1  S, x) = o”
Lfunctions. Zf x is any
{emnisI + x( l)effisi2)G(1, x)L(s, X).
We take x = h/k in Hurwitz’s formula then multiply each member by X(h) and sum on h. This gives us
PROOF.
Since F(x, s) is periodic in x with period 1 and X(h) = x(  1)x(h) we can write
hmodk
zx(1)
C X(k  h)F hmodk
and the previous formula becomes hilx(h)(’
= * (penis” (274
+ x(  l)erris’2} i
x(h)F(i,
S).
h=l
Now we multiply both members by k”’ and use (15) to obtain (16).
0 263
12 : The functions
c(s) and I.@, x)
12.11 Evaluation of c(n, a) The value of c(  n, a) can be calculated explicitly if n is a nonnegative integer. Taking s = n in the relation [(s, a) = l(1  s)Z(s,a) we find a) = l(1 + n)Z(n,
c(n,
a) = n!l(n,
a).
We also have Zn
lebz
I( n, a) = Res
z=o ( 1e’
>.
The calculation of this residue leads to an interesting class of functions known as Bernoulli polynomials. Definition
For any complex x we define the functions B,(x) by the equation B,(x) ezzeX* = cm Jz”,  1 n=O
where 1.~1< 271.
The numbers B,(O) are called Bernoulli numbers and are denoted by B,. Thus, =
z
where lzl c 2n.
ez  1
Theorem
12.12 Thefunctions
B,(x) are polynomials B,(x) = i k=O
in x given by
; Bkx”k. 0
PROOF.We have
cmB,(x) p”
n=O n.
Equating coefficients of z” we find y=kgo$& from which the theorem follows. Theorem
(17) 264
12.13 For eoery integer n 2 0 we have
c(n,u)
= y.
B,+ 164 n+l
12.12: Properties
PROOF.
of Bernoulli
numbers
and Bernoulli
polynomials
As noted earlier, we have [(  n, a) = n ! I(  n, a). Now
Cl
from which we obtain (17).
12.12 Properties of Bernoulli Bernoulli polynomials Theorem 12.14 The Bernoulli
numbers and
polynomials
B,(x) satisfy the difirence
B,(x + 1)  B,(x) = nx”’
(18) Therefore
equation
ifn 2 1.
we have B,(O) = B,( 1) ijfn 2 2.
(19)
We have the identity e’“+ l)Z zz=ze ez  1 from which we find
PROOF.
exz ez  1
m B,(x+l)B,(x) c n! n=O
xz
m x” z” = “ZO 1 z”+l.
Equating coefficients of z” we obtain (18). Taking x = 0 in (18) we obtain (19). Theorem
PROOF.
12.15 Zf n 2 2 we have
This follows by taking x = 1 in Theorem 12.12 and using (19).
0
Theorem 12.15gives a recursion formula for computing Bernoulli numbers. The definition gives B, = 1, and Theorem 12.15 yields in succession the values 1 B. = 1, B, = f, B2=;, B, = 0, B‘$= 30’ B, = 0,
8, = 0,
ho
5 = 66'
B*= Bll
30,
1
Bg = 0,
= 0. 265
12: The functions
c(s) and L(s, x)
From a knowledge of the B, we can compute the polynomials B,(x) by using Theorem 12.12. The first few are: B,(x) = x  ;,
B,(x) = 1, B&)
= x3  ; x2 + ; x,
B&)=x2x+;,
B,(x)=x42x3+x2&.
We observe that Theorems 12.12 and 12.15 can be written symbolically as follows : B,(x) = (B + x)“, B, = (B + 1)“. In these symbolic formulas the right members are to be expanded by the binomial theorem, then each power Bk is to be replaced by Bk. Theorem
12.16 Zf n 2 0 we have
Also, ijn 2 1 we have i(  2n) = 0, hence B2,+ 1 = 0.
To evaluate c(  n) we simply take a = 1 in Theorem 12.13. We have already noted that the functional equation
PROOF.
((1  s) = 2(27c“r(s)cos 09 C(s)
(21)
implies c(  2n) = 0 for n 2 1, hence B,,, 1 = 0 by (20).
0
Note. The result B2”+ 1 = 0 also follows by noting that the left member of
is an even function of z. Theorem
12.17 Zf k is n positioe integer we have Q2k) = ( l)k+ ’ (2;;;;;k
(22) PROOF.
We take s = 2k in the functional equation for c(s)to obtain c(l  2k) = 2(271) 2kI(2k)cos(zk)l(2k),
or B
 $ This implies (22).
= 2(27r) 2k(2k  l)! ( l)k[(2k). q
12.12 : Properties
of Bernoulli
numbers
and Bernoulli
polynomials
Note. If we put s = 2k + 1 in (21) both members vanish and we get no information about [(2k + 1). As yet no simple formula analogous to (22) is known for [(2k + 1) or even for any special case such as c(3). It is not even known whether [(2k + 1) is rational or irrational for any k. Theorem 12.18 The Bernoulli numbers B,, alternate in sign. That is, ( l)k+lBZk > 0. Moreover,
)B,,J + co as k + CD. Infact
( l)k+lBzk  $$
(23)
ask+co
PROOF. Since [(2k) > 0, (22) shows that the numbers B,, alternate in sign. The asymptotic relation (23) follows from the fact that [(2k) + 1 as k + 00. Note. From (23) it follows that 1B2k+2/B2kJ  k2/rr2 as k , CO. Also, by invoking Stirling’s formula, n !  (n/ey$& we find (l)k+1B2k
ask,
 47c&
oo.
The next theorem gives the Fourier expansion of the polynomial B,(x) in the interval 0 < x < 1. Theorem 12.19 ZfO < x I 1 we have (24) and hence 2(2n)! B2nC4
=
(
I).+’
02”
; k
m cos 2nkx k2” ’ 1
B PROOF. Equation (24) follows at once by taking s = n in Hurwitz’s formula and applying Theorem 12.13. The other two formulas are special cases of (24). Note. The function B,(x) defined for all real x by the right member of (24) is called the nth Bernoulli periodic function. It is periodic with period 1 and agrees with the Bernoulli polynomial B,(x) in the interval 0 < x I 1. Thus we have B,(x) = B,(x  [xl).
267
12: The functions
c(s) and L(s, x)
12.13 Formulas
for L(0, x)
Theorem 12.13 implies i(O, a) = B,(a)
= ;  a.
In particular c(O) = [(O, 1) =  l/2. We can also calculate L(0, x) for every Dirichlet character x. Theorem 12.20 Let x be any Dirichlet character mod k. (a) Zfx = x1 (the principal character), then L(0, xl) = 0. (b) Zfx # x1 we have
L(O, x) =  ; i q(r). I
Moreooer, PROOF.
1
L(0, 1) = 0 iffx(  1) = 1.
If x = x1 we use the formula
us,Xl)= a4nelk(1 p7 proved for 0 > 1 in Chapter 11. This also holds for all s by analytic continuation. When s = 0 the product vanishes so L(0, xl) = 0. If x # x1 we have L(0, x) = i x(r)c r=l
Now
= x(  1) i rX(r). r=
1
Therefore if x(  1) = 1 we have Et= IrX(r) = 0.
12.14 Approximation
0
of [(s, a) by finite sums
Some applications require estimates on the rate of growth of [(a + it, a) as a function of t. These will be deduced from another representation of {(s, a) obtained from Euler’s summation formula. This relates ((s, a) to the partial sums of its series in the halfplane c > 0 and also gives an alternate way to extend [(s, a) analytically beyond the line r~ = 1. 268
12.14: Approximation
Theorem
of
c(s,a) by finite sums
12.21 For any integer N 2 0 and CT> 0 we have
We apply Euler’s summation formula (Theorem 3.1) with f(t) =
PROOF.
(t + a)s and with integers x and y to obtain
x1= s
ycnsx (n + 4
x dt JiTiy+
x t  [t] s y (t + a)S+’ dt*
Take y = N and let x + co, keeping r~ > 1. This gives us
or
This proves (25) for D > 1. If cr 2 6 > 0 the integral is dominated by j: (t + a)‘l dt so it converges uniformly for (r 2 6 and hence represents an analytic function in the halfplane 0 > 0. Therefore (25) holds for CJ> 0 by analytic continuation. Cl The integral on the right of (25) can also be written as a series. We split the integral into a sum of integrals in which [x] is constant, say [x] = n, and we obtain m
x

1
[x] xn
s
N (x +a)S+’
dx=
(x + a)s+l
U
f n=N
s o (u+n+a)“+’
du.
Therefore (25) can also be written in the form
if 0 > 0. Integration by parts leads to similar representations in successively larger halfplanes, as indicated in the next theorem. Theorem 12.22 Zf a >  1 we have
 &1as+ 194  “f&(n+;),+1I s(s + 1) m  ~ 2!
.TN J: (n + au1 uy+’ d”*
269
12 : The
functions i(s) and L(s, x)
More generally,
ifo > m, where m = 1,2,3, . . . , we have
(28) [(s, a)  .tO &
= (N:_“‘l’S
x
[(s
 ,zl ‘(’ + ‘)&‘;‘“l,:
1
a) 
+ r,
r  ‘)
5 n=O (n + a)s+r I
i
 s(s + l)...(s + m) (m + l)!
1 Xif
p+1
s0 (n + a + u)S+~+~ d”’
n=N
Integration by parts implies
PROOF.
I
u du (n+a+ur+l=2(n+a+u)Sf’+
s+l
u2
2
u2 du f (n + a + u)S+~’
soifa>Owehave ,,!N6,n+~?u~+1=k
i: (n+a:
nN
+ q
l)s+l
IF Jo1(n + fFuy+2.
n=N
But if o > 0 the first sum on the right is [(s + 1, a)  ~~zo (n + a)S’ and (26) implies (27). The result is also valid for cr >  1 by analytic continuation. By repeated integration by parts we obtain the more general representation in (28). q
12.15 Inequalities for [[(s, a)[ The formulas in the foregoing section yield upper bounds for 1((cr + it, a) 1 as a function oft. Theorem 12.23 (a) ZfS > 0 we have I&, a)  a‘[
(29)
I i(l + 6)
if0 2 1 + 6.
(b) ZjO < 6 < 1 there is a positive constant A(6), depending on 6 but not on s or a, such that (30)
l[(s,a)
 a‘[
I A(G)(tl’
(31)
l&a)
 a‘[
I A(h)Itl’+’
(32)
l&s, a)1 < A(6)Itlm+1fd
wherem=
270
1,2,3,...
if1  6 5 CJI 2and ItI 2 1, ififm
6 I CTI 6and ItI 2 1,  6 I (r s m
+ hand ItI 2 1,
12.15: Inequalities for lc(s,a)1
PROOF.For part (a) we use the defining series for ((s, a) to obtain
which implies (29). For part (b) we use the representation in (25) when 1  6 I o < 2 to obtain
N
Itl>lwehave (N + a)‘b
Is  11
s (N + a)” I (N + 1)‘.
Finally, since JsI I lcrl + It( I 2 + (tl we find
F(N + a)” < z(N
+ a)
< +&.
These give us I&, a)  u1 < 1 + ;
+ (N + 1)6 + z
$
Now take N = 1 + [It I]. Then the last three terms are O(ltl’), where the constant implied by the Osymbol depends only on 6. This proves (30). To prove (31) we use the representation in (27). This gives us
l&,u)  usl I ~~~~+q~~~~,“+~lsl{l~(s + ;ldl
tn +;y+l
+;lslls+
+ I),4  aS‘I~ II”~N(n+lq+l.
As in the proof of (30) we take N = 1 + [It I] so that N = O(ltl) and we show that each term on the right is 0( I t 1’+‘), where the constant implied 271
12 : The
functions c(s)and L(s, 1)
by the Osymbol depends only on 6. The inequalities 6 I Q I 6 imply 1  6 5 1  ~7I 1 + 6, hence N
1 “=I =, n+a)
N
0, both I c(s)I and IL(s, x)1 are dominated by ((1 + 6) so we consider only e I 1 + 6. Theorem 12.24 Let x be any Dirichlet character mod k and assume 0 < 6 < 1. Then there is a positive constant A(6), depending on 6 but not on s or k, such thatfor s = 0 + it with ItI 2 1 we have (33)
ifm6Sol
IL(s, x)1 I A(6)1ktIm+1+b
m+6,
where m =  1, 0, 1,2, . . . PROOF.
We recall the relation kl Lhx)
=
k“x
x6K r=l
272
.
Exercises for Chapter
12
If m = 1, 2, 3, . . * we use (32) to obtain kl
IL(s,x)t
I k“1
c s,; < km+6kA(6)Itlm+1+a r=l I( )I
which proves (33) for m 2 1. If m = 0 or  1 we write (34)
Since m  6 I (r 5 m + 6 we can use (30) and (31) to obtain ,+(s,;)
 (;)‘I
5 km+dA(6),t,m+1+a,
so the second sum in (34) is dominated by A(6) 1kt jm+ ’ +*. The first sum is dominated by
and this sum can also be absorbed in the estimate A(6) I kt Im+ 1+‘.
Cl
Exercises for Chapter 12 1. Letf(n)
be an arithmetical
(a) Prove that the Dirichlet
function
which is periodic
series c f(n)n’
modulo
k.
converges absolutely
for Q > 1 and that
converges for u > 0 W If CL,f(r) = 0 Prove that the Dirichlet series 1 f(n)n’ and that there is an entire function F(s) such that F(s) = c f(n)n” for CT> 0. 2. If x is real and Q > 1, let F(x, s) denote the periodic
zeta function,
27th
F(x, s) = f >. “=l If 0 < a < 1 and d > 1 prove that Hurwitz’s I(1  s) F(a, s) = ~(27[)1 s {e”i(‘“)12[(1
formula
implies
 s, a) + eni@ 1)‘2[(1  s, 1  a)}.
3. The formula in Exercise 2 can be used to extend the definition of F(a, s) over the entire splane if 0 < a < 1. Prove that F(a, s), so extended, is an entire function of s. 4. IfO}B;:;n12;f)
for m 2 1, n 2 1. Indicate
the formula
+ tl)m+l&Bm+n
the range of the index r.
1 dx s&,(x)B,,(x)Bp(x)
20. Show that if m 2 1, n 2 1 and p 2 1, we have
0
In particular, 21. Letf(n)
compute
j; B23(~) dx from this formula.
be an arithmetical
function
which is periodic
g(n) = k
mod k, and let
1 f(m)e2”im”‘k mmodk
denote the finite Fourier
coefficients
off: If
F(s) = k”
i ,=
f(r)
0 prove that
(b) If s = Q + it with Q 2 6 > 0 and 1t 1 2 0, use part (a) to show that there is a constant A(6) such that
IUs, x)1 I 44BW(ltl where B(k) is an upper bound B(k) = O(Jis log k). (c) Prove that for some constant IL(.s,x)I [Hint:
for IS(x
+ 1)‘6
In Theorem
13.15 it is shown that
A > 0 we have
4 Alogk
ifa
1 1  andO log k
I ItI 5 2.
Take N = k in part (a).]
277
13
Analytic Proof of the Prime Number Theorem
13.1 The plan of the proof The prime number theorem is equivalent to the statement (1) *c4  x where I++(X)is Chebyshev’s function,
asx+
co,
It/(x) = c N4. “5.X This chapter gives an analytic proof of (1) based on properties of the Riemann zeta function. The analytic proof is shorter than the elementary proof sketched in Chapter 4 and its principal ideas are easier to comprehend. This section outlines the main features of the proof. The function $ is a step function and it is more convenient to deal with its integral, which we denote by tjr. Thus, we consider IcllW = Jj(r)
dt.
The integral $r is a continuous piecewise linear function. We show first that the asymptotic relation
Icllc4  ; x2 asx+cc
(4
implies (1) and then prove (2). For this purpose we express It/1(x)/x2 in terms of the Riemann zeta function by means of a contour integral, l+bl(x) x2=278
1
c+coi xsl
27ti sfmi
S(S
+ 1)
ds, where c > 1.
13.2: Lemmas
The quotient [‘(s)/[(s) has a first order pole at s = 1 with residue 1. If we subtract this pole we get the formula
We let
h(s)= 1
 g
 1
s(s + 1) ( and rewrite the last equation in the form (3)
y
 ;(1
y
= & fl
= x
sl
>
j+(s)ds +m
_ h(c + it)e” ‘Ogxdt. s 03
To complete the proof we are required to show that y1
+m
h(c + it)e” “gx dt = 0. lim ~ x+m 27T s cc Now the RiemannLebesgue lemma in the theory of Fourier series states that +03 lim f(t)e”” dt = 0 xcc s 03 (4)
if the integral J? z 1f(t) 1dt converges. The integral in (4) is of this type, with x replaced by log x, and we can easily show that the integral 1’ z 1h(c + it) I dt converges if c > 1, so the integral in (4) tends to 0 as x + co. However, the factor xc ’ outside the integral tends to co when c > 1,so we are faced with an indeterminate form, co . 0. Equation (3) holds for every c > 1. If we could put c = 1 in (3) the troublesome factor xc’ would disappear. But then h(c + it) becomes h(1 + it) and the integrand involves [‘(s)/T(s) on the line CJ= 1.In this caseit is more difficult to prove that the integral 1’ 2 1h(1 + it) I dt converges, a fact which needs to be verified before we can apply the RiemannLebesgue lemma. The last and most difficult part of the proof is to show that it is possible to replace c by 1 in (3) and that the integral j? z 1h(1 + it) I dt converges. This requires a more detailed study of the Riemann zeta function in the vicinity of the line cr = 1. Now we proceed to carry out the plan outlined above. We begin with some lemmas.
13.2 Lemmas Lemma 1 For any arithmeticalfunction 44
a(n) let = c44, “5X 279
13: Analytic proof of the prime number theorem where A(x) = 0 ifx < 1. Then
hOOF.
We apply Abel’s identity (Theorem 4.2) which states that
(6)
14n)f(n) nsx
= A(x)f(x)

j)(r)fV)
dt
iffhas a continuous derivative on [I, x]. Takingf(t)
“gNf(4 =“5.x c 44
and A(x)f(x)
= t we have = x 1 a(n) n5.x
so (6) reduces to (5).
. q
The next lemma is a form of L’HGspital’s rule for increasing piecewise linear functions. Lemma 2 Let A(x) = znSx a(n) and let A,(x) = s; A(t) dt. Assume also that u(n) 2 0 for all n. If we have the asymptotic formula A,(x) ‘v Lx’ us x + c0 (7) for some c > 0 and L > 0, then we also have us x + co. A(x)  cLx’l (8) In other words, formal difirentiution of (7) gives a correct result. PROOF. The function A(x) is increasing since the u(n) are nonnegative. Choose any fi > 1 and consider the difference A,@x)  A,(x). We have
A,(/?x)  A,(x) = /“A(u) du 2 j+A(x) x x = x(/l  l)A(x).
du = A(x)@x  x)
This gives us
x4x) I !L p _ 1 {AI
 A,(41
or
44
~1
xcl s /?  1
AI B’ (/?x)
“$9).
Keep /? fixed and let x ) COin this inequality. We find 44 lim sup p&LB’L)=LE. xa, 280
13.2: Lemmas
Now let B + 1 +. The quotient on the right is the difference quotient for the derivative of xc at x = I and has the limit c. Therefore lim sup A(x) I CL.
(9)
x+m x
Now consider any u with 0 < a < 1 and consider the difference A,(x)  A,(ctx). An argument similar to the above shows that liminf$>
Ls.
xm
As CIP 1 the right member tends to CL. This, together with (9) shows that A(x)/x’’ tends to the limit CL as x + 03. Cl When a(n) = A(n) we have A(x) = $(x), A,(x) = @r(x), and a(n) 2 0. Therefore we can apply Lemmas 1 and 2 and immediately obtain: Theorem
13.1 We have
Also, the asymptotic relation til(x)
 x212 implies +(x)  x as x + 00.
Our next task is to express $r(x)/x’ as a contour integral involving the zeta function. For this we will require the special cases k = 1 and k = 2 of the following lemma on contour integrals. (Compare with Lemma 4 in Chapter 11.) Lemma 3 Zf c > 0 and u > 0, then for every integer k 2 1 we have
I,,
I
z(z + I;...(z
+ k)dz=
 U)k $0 < 2.45 1,
k! 0
if24 > 1,
the integral being absolutely convergent. PROOF. First we note that the integrand is equal to u~T(z)/T(z + k + 1).
This follows by repeated use of the functional equation T(z + 1) = zT(z). To prove the lemma we apply Cauchy’s residue theorem to the integral 1 uw) dz I2xi cCRjT(z + k + 1) ’ where C(R) is the contour shown in Figure 13.1(a) if 0 < u I 1, and that in Figure 13.1(b) if u > 1. The radius R of the circle is greater than 2k + c so all the poles at z = 0,  1, . . . ,  k lie inside the circle. 281
13 : Analytic
proof of the prime
number
theorem
// /I Ii I \ \\ \\
\ \ \ I I //
(a)
O 1. Now if 1 5 n I k we have (z + nl 2 IzI  n = R  n 2 R  k 2 R/2
since R > 2k. Therefore the integral along each circular arc is dominated by 2rtRu ’ ~ = O(Rk)
R($Rr
and this PO as R + 00 since k 2 1. If u > 1 the integrand is analytic inside C(R) hence &) = 0. Letting R + cc we find that the lemma is proved in this case. If 0 < u I 1 we evaluate the integral around C(R) by Cauchy’s residue theorem. The integrand has poles at the integers n = 0,  1, . . . , k, hence 1 2ni I
u  ‘I(z) C(R) rb
+ k + 1)
dz = i
Res
u “I(z)
“=,, =z,, I(z + k + 1)
= “$ r(k +“;  n) En
Letting R + co we obtain the lemma. 282
‘(‘) = ,jo (;‘,;H,
!
cl
A contour
13.3:
13.3 A contour for
Theorem
(11)
integral
integral
representation
for t,b1(x)/x2
representation
rl/ 1(4/x”
13.2 Zfc > 1 and x 2 1 we have
l+bl(X) X2
1 c+mi xsi = ~ scmi S(S+ 1)
PROOF.From Equation (10) we have $i(x)/x = xnsx (1  n/x)A(n). Now use Lemma 3 with k = 1 and u = n/x. If n I x we obtain c+mi W)s
lr,i X
Multiplying y
ds.
27ri scooi s(s + 1)
this relation by A(n) and summing over all n < x we find = xx & J;Iii
c+mi A(n)(x/n) ds = jf, & lemi s(s + 1) ds
;;:‘I;’
since the integral vanishes if n > x. This can be written as (12)
y
= fl
[+R’fn(s)
ds,
cmi
where 27$“(x) = A(n)(x/n)“/(s2 + s).Next we wish to interchange the sum and integral in (12). For this it suffices to prove that the series (13)
“El [+a’lSn(~)I
ds
cmi
is convergent. (See Theorem 10.26 in [2].) The partial sums of this series satisfy the inequality
where A is a constant, so (13) converges. Hence we can interchange the sum and integral in (12) to obtain
Now divide by x to obtain (11).
cl 283
13 : Analytic proof of the prime number theorem
Theorem 13.3 Ifc
z=1 and x 2 1 we have
(14) where
PROOF.
This time we use Lemma 3 with k = 2 to get f(1~~=21n;~~~~s~s+;;;9+2)dS1
where c > 0. Replace s by s  1 in the integral (keeping c > 1) and subtract the result from (11) to obtain Theorem 13.3. q
X
sl
If we parameterize the path of integration by writing s = c + it, we find _ xclxit = xcleit logx and Equation (14) becomes
Our next task is to show that the right member of (16) tends to 0 as x + co. As mentioned earlier, we first show that we can put c = 1 in (16). For this purpose we need to study C(s)in the neighborhood of the line CJ= 1.
13.4 Upper bounds for (c(s)/ and /[‘@)I near the line CJ= 1 To study C(s)near the line cr = 1 we use the representation obtained from Theorem 12.21 which is valid for e > 0, (17)
i(s)= “Cl;
m S s N
x

[x]
XI+ldx+.
NlS
sl
We also use the formula for c’(s) obtained by differentiating of (1%
(18) &)= 5 logn+s~(x~~~gxdx~~x~,l:l,x n=l ns N
 N’“log
sl
N
each member
N
N’”
(s
The next theorem uses these relations to obtain upper bounds for 1c(s)1 and I r’(s) I. 284
13.4: Upper
bounds for 1i(s) 1and I[‘(s)l
near the line o = 1
Theorem 13.4 For every A > 0 there exists a constant M (depending on A) such that Il
A log t
and c 2 e.
Note. The inequalities (20) describe a region of the type shown in Figure 13.2.
0
1
0
Figure
13.2
PROOF. If CJ2 2 we have 1&)I I c(2) and 1T’(s)1< l[‘(Z)l and the inequalities in (19) are trivially satisfied. Therefore we can assume Q < 2 and t 2 e. We then have
IslIa+tI2+t 1).
This can be written as
from which we find 1i(s) 1= exp 1 f cos(Ii2g ‘)}. i p m=l We apply this formula repeatedly with s = 0, s = 0 + it and s = 0 + 2it, and obtain
13(4 Iito + it) I4I ita + 24 I m 3 + 4 cos(mt log p) + cos(2mt log p) mpmo i p m=l
= exp C C
But we have the trigonometric inequality 3 + 4cose+ 286
cos202
0
13.6:
Inequalities for Il/c(s) l and l [‘(s)/[(s) l
which follows from the identity 3 + 4 cos8 + cos28 = 3 + 4 cose + 2 ~0~28  i = 2(1 + cose)z. Therefore each term in the last infinite series is nonnegative so we obtain
q
(21). Theorem
13.6 We haue ((1 + it) # Ofor euery real t.
PROOF. We need only consider t # 0. Rewrite (21) in the form (22)
{(CT l)&r)}3
4,1(fJ + 2it)l 2 &.
3 I
I
This is valid if 0 > 1. Now let (r + 1+ in (22). The first factor approaches 1 since i(s) has residue 1 at the poles = 1.The third factor tends to l C(1 + 2it)l. If [(l + it) were equal to 0 the middle factor could be written as [(a + it)  ((1 + it) 4 + 1[‘(l + it) l4 as c + 1 + . Ol Therefore, if for some t # 0 we had [(l + it) = 0 the left member of (22) would approach the limit l c(l + it) I4 lc( 1 + 2it) l as 0 + 1 + . But the right member tends to cc as c + 1+ and this gives a contradiction. 0
13.6 Inequalities
for [l/[(s)/
and I~‘(s)/~(s)I
Now we apply Theorem 13.5 once more to obtain the following inequalities
for I l/i(s) I and Ii’bM4 I. Theorem
13.7 There is a constant M > 0 such that
ll
1 < M log’ a.4
t
5’(s)
and 50
I
I
< M log9 t
whenever c 2 1 and t 2 e.
PROOF.For o 2 2 we have
and
so the inequalities hold trivially if o 2 2. Suppose, then, that 1 < 0 < 2 and t 2 e. Rewrite inequality (21) as follows: 1
,r(a + it), I G43’4 ILX0+ 2it) 11’4. 287
13 : Analytic
proof of the prime number
theorem
Now (0  l)[(o) is bounded in the interval 1 < r~ < 2, say (a  l){(g) 5 M, where M is an absolute constant. Then a4
5 ~
M
a1
if 1 < 0 I 2.
Also, ((a + 2it) = O(log t) if 1 5 (r < 2 (by Theorem 13.4), so for 1 < 0 5 2 we have 1 1[(a + it)1 ’
M3’4(log t)l’4 A(log t)“4 (0  1)3/4 = (a  1)3/4 ’
where A is an absolute constant. Therefore for some constant B > 0 we have (23)
I5(o + it)1 >
B(o  1)3’4 (log t)I,4 , if 1 < CJI 2, and t 2 e.
This also holds trivially for G = 1. Let CYbe any number satisfying 1 < LX< 2. Then if 1 I Q I a, t 2 e, we may use Theorem 13.4 to write dlljl(t4+it)ldt4~(cxa)Mlog2t sd
15(0+it)C(a+it)lI
I (a  l)M log2 t. Hence, by the triangle inequality, I [(a + it)1 2 I [(a + it)1  I i(a + it)  [(a + it)1 2 1[(a + it)1  (a  l)M log2 t 2
B(a  1)3’4 (log t)I,4  (a  l)M log2 t.
This holds if 1 5 (r I a, and by (23) it also holds for a I cr I 2 since (a  1)3’4 >(a  1)3’4. In other words, if 1 I c I 2 and t 2 e we have the inequality B(a  1)3’4
Ii@ + 91 2 tlog tj1,4  (a  l&f log2 t for any a satisfying 1 < a < 2. Now we make a depend on t and choose a so the first term on the right is twice the second. This requires
Clearly a > 1 and also a c 2 if t 2 t, for some to. Thus, if t 2 t,, and 1 I 0 I 2 we have IC(o + it)1 2 (a 
C 1)M log2 t = (log t17.
The inequality also holds with (perhaps) a different C if e < t s to. 288
13.7 : Completion
of the proof of the prime number
theorem
This proves that 1c(s)1 2 C log’ t for all o 2 1, t 2 e, giving us a corresponding upper bound for 1l/&)1. To get the inequality for I &)/&)I we apply Theorem 13.4 and obtain an extra factor log’ t. 0
13.7 Completion of the proof of the prime number theorem Now we are almost ready to complete the proof of the prime number theorem. We need one more fact from complex function theory which we state as a lemma. Lemma 4 Zff(s) has a pole of order k at s = a then the quotientp(s)/f(s) jirst order pole at s = a with residue k.
has a
PROOF.We have f(s) = g(s)/(s  a)k, where g is analytic at a and g(a) # 0. Hence for all s in a neighborhood of a we have
g’(s) f’(‘)
=
(s
a)k
kg(s) ts
_
a)k+
1
Thus f’(s) =f(s)
k sa
I d(s) g(s)
This proves the lemma since g’(s)/&) is analytic at a.
0
Theorem 13.8 Thefunction qs) = r’(s)  1 Sl i(s) is analytic at s = 1.
PROOF.By Lemma 4, c’(s)/[(s) has a first order pole at 1 with residue 1, as does l/(s  1). Hence their difference is analytic at s = 1. cl Theorem
13.9 For x 2 1 we have
where the integral f? o. I h( 1 + it) I dt converges. Therefore, Lebesgue lemma we have
by the Riemann
Ii/l(X)  x212
(24) and hence be)
N x
asx,
co. 289
13: Analytic proof of the prime number theorem
PROOF.
In Theorem
13.3 we proved that if c > 1 and x 2 1 we have
where
h(s)= A
s(s I 1) (
_ r’(s)_ 1
sl
Us)
>*
Our first task is to show that we can move the path of integration to the line g = 1. To do this we apply Cauchy’s theorem to the rectangle R shown in Figure 13.3. The integral of xS l h(s) around R is 0 since the integrand
‘t
. 
.
I +iT
c+iT
1
*LJ
I
0
c
1
I
I

I
I
1 iT
ciT
Figure 13.3
is analytic inside and on R. Now we show that the integrals along the horizontal segments tend to 0 as T + co. Since the integrand has the same absolute value at conjugate points, it suffices to consider only the upper segment, t = 7: On this segment we have the estimates
I I S(Sq1
5F 1
and
1 1 1 s(s + l)(s  1) S Tj S T2.
Also, there is a constant M such that / c(s)/ 0 and C > 0 such that
whenever (25)
l
A log9
0. To estimate the last term we write li’(u + it)du C ‘I&4 + it)1 du. Is I7 I J0 Since t 2 e we have log9 t 2 log; t so 1  (A/log9 t) 2 1  (A/log t). Thus, if g satisfies (25) for any A > 0 we can apply Theorem 13.4 to estimate )5’(u + it) (, giving us MA li(l+it)i(r~+it)J<M(lo)logZ~<Mlog21&= log7. Using this in (26) we find I[(1 + it)  ((0 + it)I =
I((0 + it)1 > “,,y
.
This holds for some B > 0, any A > 0 and some M > 0 depending on A. A value of M that works for some A also works for every smaller A. Therefore we can choose A small enough so that B  MA > 0. If we let C = B  MA the last inequality becomes )c(a t it) I > C log ’ t which proves the theorem for all 0 and t satisfying A l 0 we have (1  p’s)Qs)
= f q2. PI=1
This implies that C(s)< 0 ifs is real and 0 < s < 1. 292
13.9: The Riemann
PROOF.
hypothesis
First assume that 0 > 1. Then we have
= (1 + 2” + 3” + . . .)  2(2” + 4” + 6” + . . .) = 1 _ 2s + 3s  4s + 5s _ 6s + . . . , which proves (27) for rr > 1. However, if Q > 0 the series on the right converges, so (27) also holds for cr > 0 by analytic continuation. When s is real the series in (27) is an alternating series with a positive sum. If 0 < s < 1 the factor (1  2l“) is negative hence c(s) is also negative. 0
13.9 The Riemann hypothesis In his famous &page memoir on rr(x) published in 1859, Riemann [58] stated that it seems likely that the nontrivial zeros of c(s) all lie on the line r~ = l/2, although he could not prove this. The assertion that all the nontrivial zeros have real part l/2 is now called the Riemann hypothesis. In 1900 Hilbert listed the problem of proving or disproving the Riemann hypothesis as one of the most important problems confronting twentieth century mathematicians. To this day it remains unsolved. The Riemann hypothesis has attracted the attention of many eminent mathematicians and a great deal has been discovered about the distribution of the zeros of C(s).The functional equation shows that all the nontrivial zeros (if any exist) must lie in the strip 0 < g < 1, the socalled “critical strip.” It is easy to show that the zeros are symmetrically located about the real axis and about the “critical line” CJ= l/2. In 1915 Hardy proved that an infinite number of zeros are located on the critical line. In 1921 Hardy and Littlewood showed that the number of zeros on the line segment joining l/2 to (l/2) + iT is at least AT for some positive constant A, if T is sufficiently large. In 1942 Selberg improved this by showing that the number is at least AT log T for some A > 0. It is also known that the number in the critical strip with 0 < t < T is asymptotic to T log T/2n: as T + co, so Selberg’s result shows that a positive fraction of the zeros lie on the critical line. Recently (1974) Levinson showed that this fraction is at least 7/10. That is, the constant in Selberg’s theorem satisfies A 2 7/2Orc. Extensive calculations by Gram, Backlund, Lehmer, Haselgrove, Rosser, Yohe, Schoenfeld, and others have shown that the first threeandahalf million zeros above the real axis are on the critical line. In spite of all this evidence in favor of the Riemann hypothesis, the calculations also reveal certain phenomena which suggest that counterexamples to the Riemann hypothesis might very well exist. For a fascinating account of the story of largescale calculations concerning c(s) the reader should consult [17]. 293
13: Analytic proof of the prime number theorem
13.10 Application to the divisor function The prime number theorem can sometimes be used to estimate the order of magnitude of multiplicative arithmetical functions. In this section we use it to derive inequalities for d(n), the number of divisors of n. In Chapter 3 we proved that the average order of d(n) is log n. When n is prime we have d(n) = 2 so the growth of d(n) is most pronounced when n has many divisors. Suppose n is the product of all the primes IX, say n = 2,3.5.... (28) Pn(x). Since d(n) is multiplicative
we have
d(n) = d(2)d(3) . . . d(p,,,,) = 2’+). For large x, rc(x) is approximately x/log x and (28) implies that log n = 1 log p = 9(x)  x PSX so 2ncx)is approximately 2 log”/ loglog”. Now 2a 1ogn _ ea logn log2 = nn log2 2 l%n/log logn= n log2/ loglogn. In other words, when n is of the form (28)
hence
then
d(n)
is approximately
2 log “/ log ‘a ” =
n log 2/ log log n.
By pursuing this idea with a little more care we obtain the following inequalities for d(n). Theorem
13.12 Let
E
> 0 be given. Then we have:
(a) There exists an integer N(E) such that n 2 N(E) implies (qn)
2” &) log n/
log
log”
=
log n =
,(l
+ E) log
2/ log
log n.
log
2/ log
log n*
,(1E)
Note. These inequalities are equivalent to the relation log d(n)log log n = log 2. lim sup log n n+m PRooF. Write n = pie* . . . pka*, so that d(n) = nf=i (ai + 1). We split the product into two parts, separat.ing those prime divisors < f(n) from those 2 f(n), wheref(n) will be specified later. Then d(n) = Pl(n)P2(n) where PI(n)
294
=
lI
(ai
Pi< f(n)
+
1)
and P2(n) =
n Pi 2 f(n)
(ai + 1).
Application to the divisor function
13.10:
In the product P,(n) we use the inequality (a + 1) I 2” to obtain P,(n) I 2’(“), where S(n) =
~
Ui.
i=l Pi 1 f(n)
Now n = fip,“’ i=l
2
fl
pFi 2
Pi z f(n)
n
f(nyi = f(n)s(“),
Pi 2 J(n)
hence
log n or S(n) I log f(n).
log n 2 S(n)log f(n), This gives us
p,(n) 5 2 losn/logf(n). (29) To estimate Pi(n) we write PI(n) = exp
1 lOg(Ui + 1) r Pi 1 define
(a) Derive the following
finite identities
F
m=l
of Shanks:
X*0 1)/Z= El ~Q.(x) XS(2n+ s=o QsC4
1) ’
2n+l
(b) Use Shanks’ identities
to deduce Gauss’ triangularnumber
theorem:
for 1x1 < 1.
f
m=l
6. The following
identity
is valid for 1x 1 < 1: X*(*+ U/2 = fJ (1 + Xn I)(1  x2y,
f *=LX
n=1
(a) Derive this from the identities in Exercises 2 and 5(b). (b) Derive this from Jacobi’s triple product identity. 7. Prove that the following triple product identity: (a) “fil(l
 ,$“)(l
(b) fi(1
 x5”)(1 
identities,
valid for 1x 1 < 1, are consequences
 x5n4) =
 f+l)(l
x5n2)(1
__ X5n3)
=
n=l
of Jacobi’s
f (l)mX*(5*+3)12. *=m $
(l)*X*(~*+1K2~
*=LX
8. Prove that the recursion
formula v(n)
= i
4kMn
 4,
k=l
obtained
in Section 14.10, can be put in the form q(n)
=
i
1
m=l
ksnlm
mp(n  km).
9. Suppose that each positive integer k is written in g(k) different colors, where g(k) is a positive integer. Let p,(n) denote the number of partitions of n in which each part k appears in at most g(k) different colors. When g(k) = 1 for all k this is the
326
Exercises for Chapter
unrestricted partition function p(n). Find p,(n) and prove that there is an arithmetical
an infinite product functionf(depending
14
which generates on g) such that
v,(n) = i f(k)p,(n  4. k=l
10. Refer to Section 14.10 for notation. By solving in (22) prove that if 1x 1 < 1 we have “p
 ym/n
the firstorder
differential
equation
= exp{[Iq)dt},
where
ff(x) = 2 fXdxk and f2k) = C f(4 k=l Deduce that fj(1 where p(n) is the Mobius The following
a proof of Ramanujan’s
+ 4)x” = 5 $f$,
of Kruyswijk
for 1x1 < 1,
function.
exercises outline
m$~(5m by a method
 x”)~(“)~” = eex
partition
identity
where q(x) = fi (1  x”), n=1
not requiring
the theory of modular
functions,
11. (a) Let E = eZniik where k 2 1 and show that for all x we have *fil(l (b) More
generally,
 X&h) = 1  xk.
if (n, k) = d prove that *f!(l
 X&“h) = (1  Xk’d)d,
and deduce that if(n,k)=
*fil(l  Xne2nWk)  {ilz’)k 12. (a) Use Exercise 1 l(b) to prove that for prime
“I1 *fil(l
1,
ifkln.
q and Ix 1 < 1 we have
 x”eZninhiq)= s.
(b) Deduce the identity mfj,p(rn)xm
= $$
*fJl “fir(l
 xneZainh’5).
327
14: Partitions
13. If q is prime and if 0 2 I < q, a power series of the form
is said to be of type r mod q. (a) Use Euler’s pentagonal number power series,
theorem
q(x) =: fi(l “=I
to show that q(x) is a sum of three
 x”) = I, + I, + I,,
where I, denotes a power series of type k mod 5. (b) Let tl = ezni15 and show that
x”a”h) = fi(r, + I,ah+ I,CrZh). h=l
(c) Use Exercise 12(b) to show that fop(5m
+ 4)X5m+4 = v, sj,
where V, is the power series of type 4 mod 5 obtained 14. (a) Use Theorem power series,
from the product in part(b).
14.7 to show that the cube of Euler’s product
is the sum of three
q’(x)3 = w, + w, + w,, where W, denotes a power series of type k mod 5. (b) Use the identity W, + W, + W, = (I, + I, + 1J3 to show that the power series in Exercise 13(a) satisfy the relation I,12
= I,Z.
(c) Prove that I, = x(p(xz5). 15. Observe that the product flz= r (I, + 1,~~ + 1,~‘~) is a homogeneous polynomial in I,, I,, I, of degree 4, so the terms contributing to series of type 4 mod 5 come from the terms 114, 1,1,212 and I,’ I,‘. (a) Use Exercise 14(c) to show that there exists a constant c such that v, = cI,4, where V, is the power series in Exercise 13(c), and deduce that
m;/hl + 4)xSm+4 = cx4$g. (b) Prove that c = 5 and deduce Ramanujan’s m$(5m
328
identity
+ 4)xm = 5 ‘PO5 (Pc46
Bibliography
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332
Index of Special Symbols
d In, d Y n, (4 b), h,
. . . 3 4,
[a, bl, An), cp(n), f * 99
[I
divides (does not divide), 14 greatest common divisor (gcd),l5,21 least common multiple (lcm), 22 Mobius function, 24 Euler totient, 25 Dirichlet convolution, 29
Z(n) = i ,
identity function, 30
f5
Dirichlet inverse, 30 unit function, 31 Mangoldt function, 32 Liouville function, 37 divisor functions, 38 generalized convolution, 39 Bell series off modulo p, 43 derivative, 45 Euler’s constant, 53 big oh notation, 53 asymptotic equality, 53 Riemann zeta function, 55 number of primes IX, 74 Chebyshev +function, 75
u(n) =
1,
A(n), W, o,(n), 44, d(n), a 0 F,
fpM
f ‘(4 = f (n)logn, C, 0, i;s;, 4x), $64
333
Index of special symbols
%4> M(x), 0, a = b (mod m),
^ a, I a, x(n), L(L xl, L’U, x), 44, G(n, xl, W ; 4, nRp, nb, b IPX (n I PI, evmk4 ind, a, Lb, xl, 0a, fJ ri;,,
as, 4, F(x, 4, 8k4
4,
m4 p(n), dn),4
334
 4,
Chebyshev &function, 75 partial sums of Mobius function, 91 little oh notation, 94 congruence, 106 residue class a module m, 109 reciprocal of a modulo m, 111 Dirichlet character, 138 sum of series C x(n)/n, 141 sum of series c X(n)log n/n, 148 Ramanujan sum, 160 Gauss sum associated with x, 165 quadratic Gauss sum, 177 quadratic residue (nonresidue) mod p, 178 Legendre symbol, 179 Jacobi symbol, 188 exponent of a module m, 204 index of a to base g, 213 Dirichlet Lfunction, 224 abscissa of absolute convergence, 225 abscissa of convergence, 233 gamma function, 250 Hurwitz zeta function, 251 periodic zeta function, 257 Bernoulli polynomials, (numbers), 264 periodic Bernoulli functions, 267 partition function, 307 pentagonal numbers, 311
Index
Abel, Niels Henrik, 77 Abelian group, 129 Abel’s identity, 77 Abscissa, of absolute convergence, 225 of convergence. 233 Additive number theory, 304 Algorithm, division, 19 Euclidean, 20 Analytic continuation, of Dirichlet Lfunctions, 255 of Hurwitz zeta function, 254 of Riemann zeta function, 255 Apostol, Tom M., 329 Arithmetic, fundamental theorem of, 17 Arithmetical function, 24 Asymptotic equality, 53 Average (arithmetic mean) of an arithmetical function, 52 Average order, 52 Ayoub, Raymond G., 329 Bell, Eric Temple, 29, 42, 329 Bell series, 43 Bernoulli, numbers, 265 periodic functions, 267 polynomials, 264 Binomial congruence, 214 Borozdkin, K. G., 10, 329 BuhStab, A. A., 11,329 Cauchy, AugustinLouis, 44, 144, 198 Cauchy product, 44 Chandrasekharan, Komaravolu, 329 Character, Dirichlet, 138 of an abelian group, 133 primitive, 168 principal, 138
Chebyshev, Pafnuti Liwowich, 9. 75 Chebyshev function 9(x), 75 Chebvshev function tilx). 75 Chen: Jingrun, I 1, 3b4,329 Chinese remainder theorem, I 17 Clarkson, James A., 18,330 Classes of residues, 109 Clausen, Thomas, 275 Common divisor, 14 Commutative group, 129 Complete residue system, 110 Completely multiplicative function, 33 Conductor of a character, 17 1 Congruence, 106 Convolution. Dirichlet. 29 generalized, 39 van der Corput, J. G., 59 Critical line. 293 Critical strip, 293 Crossclassification principle, 123 Cyclic group, 131 Darling, H. B. C., 324 Davenport, Harold, 330 Decomposition property of reduced residue systems, 125 Derivative of arithmetical functions, 45 Dickson, Leonard Eugene, 12, 330 Diophantine equation, 5, 190 Dirichlet, Peter Gustav Lejeune, 5, 7, 29, 53, 138. 146,224 Dirichlet, character I. 138 convolution (product). 29 divisor problem, 59 estimate for d(n), 53, 57 inverse, 30 Lfunction, 224 series, 224 theorem on primes in arithmetic progressions, 7, 146, 154 Disquisitiones arithmeticae, 5 Divisibilitv, 14 Division algorithm, 19 Divisor, 14 Divisor function e.(n), 38 Edwards, H. M., 330 Elementary proof of prime number theorem, 9.98 Ellison, W. J., 306, 330 Erdiis, Paul, 9, 330 Euclidean algorithm, 20 Euclid’s lemma, 16 Euler, Leonhard, 4, 5, 7, 9, 19, 25, 53, 54, 113, 180,185,230,308, 312,315 EulerFermat theorem, 113 Euler product, 230 Euler’s constant, 53, 250 Euler’s criterion, 180
335
Index
Euler’s pentagonalnumber theorem, 3 12 Euler’s summation formula, 54 Euler totient function (p(a). 25 Evaluation, of (  11p), 181 of(2lp), 181 of [( n, a), 264 of 1(2n), 266 of L(0, x), 268 Exponent of a module m, 204 Exponential congruence, 215 Factor, 14 Fermat, Pierre de, 5, 7, 11, 113, 114 Fermat conjecture, I I Fermat prime, 7, Fermat theorem (little), 114 Finite Fourier expansion, 160 Formal power series, 41 Fourier coefficient, 160 Franklin, Fabian, 3 13, 330 Function, arithmetical, 24 Bernoulli neriodic B.(x). 267 Chebyshev 9(x), 75 “’ Chebyshev I/I(Y), 75 completely multiplicative, 33 Dirichlet L(s, x), 224 divisor d(n), u,(n), 38 Euler totient cp(n), 25 Hurwitz zeta ((s, a), 249 Liouville I(n), 37 Mangoldt A(n). 32 Mobius &I), 24 periodic zeta F(x, s), 257 Riemann zeta c(s), 9, 249 K(n), 247 M(x), 9 1 v(n), 247 n(x), 8.74 Ii/,(x), 278 Functional equation, for F(s), 250 for L(s, x), 263 for i(s), 259 for i(s, h/k), 261 Fundamental theorem of arithmetic, 17 Gamma function, 250 Gauss, Carl Friedrich. 5, 7, 8, 106, 165, 177, 182, 185, 306, 326 Gauss sum, associated with x, 165 quadratic, 177, 306 Gauss’ lemma, 182 Gauss’ triangularnumber theorem, 326 Generating function, 308 Geometric sum, 157, 158 Gerstenhaber, Murray, 186, 330 Goldbach, C., 6,9, 304 Goldbach conjecture, 9, 304 Greatest common divisor, 15,20, 21 Greatest integer symbol, 8,25, 54, 72 Grosswald, Emil, 330
336
Group, definition abelian. 129 cyclic, I3 I Group character.
of. 129
133
Hadamard, Jacques, 9,74, 330 Hagis, Peter, Jr., 5, 330 Halfplane, of absolute convergence, 225 of convergence, 233 Hardy, Godfrey Harold, 59,293, 305, 330 Hemer, Ove, 330 Hilbert, David, 293 Hurwitz, Adolf, 249 Hurwitz formula for {(s, u), 257 Hurwitz zeta function [(s, a), 249, 251, 253, 255 Identity element. 30, 129 Identity function I(n), 30 Index, 213 Index calculus, 214 Indices (table of),216, 217 Induced modulus, 167 Induction, principle of, 13 Infinitude of primes, 16, 19 Inequalities, for 1[(s, a) 1, 270 for 1i(s) 1, 270, 287, 291 for I Us, x) I, 272 for n(n), 82 for nth prime p.. 84 for d(n), 294 for q(n), 298 for p(n), 316, 318 Ingham, A. E., 330 Inverse, Dirichlet, 30 of completely multiplicative function, Inversion formula, Mobius, 32 generalized, 40 Iseki, Kaneshiro, 99, 332 Jacobi, Jacobi Jacobi Jordan
36
Carl Gustav Jacob, 187,305, 313, 319 symbol (n I P), 188 triple product identity, 319 totient Jk(n), 48
Kloosterman, H. D., 176 Kloosterman sum, 176 Kolberg, Oddmund, 324 Kolesnik. G. A.. 59 Kruyswijk, D., 324, 327, 331 Lagrange, Joseph Louis, 5, 115, 144, 158 Lagrange interpolation formula, 158 Lagrange’s theorem on polynomial congruences, 115 Landau, Edmund, 59,237,248,301,331 Landau’s theorem, 237,248 Lattice points, 57, 62 visibility of, 62
Index
Law of quadratic reciprocity. 185. 189, 193.200 Least common multiple, 22 Leech. John. 10.331 Legendre, AdrienMarie, 5, 67, 179, 185, 331 Legendre’s identity, 67 Legendre symbol (n Ip), 179 Lehmer, Derrick Henry, 6, 293, 316, 331 Lehmer, Derrick Norman, 6, 331 Lemma of Gauss, 182 LeVeque, William Judson, 2, 191, 33 1 Levinson, Norman, 293, 33 1 Lfunction L(s, x), 224 Linear congruence, 111, 112, 114,214 van Lint, Jacobus Hendricus, 318, 331 Liouville, Joseph, 37 Liouville function A(n), 37 Little Fermat theorem, I14 Littlewood, John Edensor, 10, 305, 331 Logarithmic integral Li(x), 102 Lucas, Edouard, 275 MacMahon, Percy A., 316 von Mangoldt, H., 32 von Mangoldt function A(n), 32 Mean value formulas for Dirichlet series, 240 Mersenne, P., 4 Mersenne numbers, 4 Mertens, Franz, 91 Mertens’ conjecture, 9 1 Mills, W. H., 8, 331 Mobius, Augustus Ferdinand, 24 Mobius function p(n), 24 Mobius function pLk(n) of order k, 50 Mobius inversion formula, 32 product form, 47 generalized, 40 Mordell, Louis Joel. 305. 306 Multiplication, Dirichlet. 29 of residue classes, 138 Multiplicative function, 33 Multiplicative number theory, 304 Nevanlinna, Veikko, 331 Niven, Ivan, 331 Nonresidue, 178 Numbertheoretic function,
24
0, big oh notation, 53 o, little oh notation, 94 Order of a group, 130 Orthogonality relation, for group characters, 137 for Dirichlet characters, 140 Partition, 304 Partition function p(n), 307 Pentagonal numbers, 2, 5, 311 Pentagonalnumber theorem, 312 Perfect numbers, 4
Periodic arithmetical function, 157 Periodic zeta function, 257 Polya, G., 173, 299 Polya inequality for character sums. 173. 176. 299 Polygonal numbers, 2, 5 Polynomial congruence, 115 Prachar, Karl, 331 Prime number theorem, 9,65,74, 79,92,94, 98,278,289 Primes, 2, 16, contained in a factorial, 67 in arithmetic progressions, 7, 146, 154 Fermat, 7 infinitude of, 16, 19 Mersenne, 4 Primitive character, 168 Primitive root, 204 Principal character, 134, 138 Product, of arithmetical functions, 29 of Dirichlet series, 228 Pythagorean triple, 2 Quadratic, congruence, 178 Gauss sum, 177, 195 nonresidue, 178 reciprocity law, 185, 189, 193, 200 residue. 178 Rademacher, Hans, 3 16, 33 1 Ramanuian. Srinivasa. 160. 305. 324. 328 Ramanujan’partition identities, 324,‘328 Ramanujan sum, 160, 176 Reciprocity law, for Jacobi symbols, 189 for Legendre symbols, 185, 193,200 for quadratic Gauss sums, 200 Reduced fraction, 21 Relatively prime, 15, 21 in pairs, 21 Rinyi, Alfred, 10, 332 Residue, quadratic, 178 Residue class, 109 Residue system, complete, 110 reduced, 113, 125 Riemann, Georg Friedrich Bernhard, 9,225 293,332 Riemann hypothesis,,293, 301 RiemannLebesgue lemma, 279 RiemannStieltjes integral, 77 Riemann zeta function, 249 Ring of formal power series, 42 Robinson, Raphael M., 7, 332 Rosser, J. Barkley, 293, 332 Schnirelmann, L., 10, 332 Selberg. Atle, 9. 46. 100. 332 Selberg asymptotic formula, 100, 103, 104 Selberg identity, 46 Separable Gauss sums, 165, 171 Shanks, Daniel, 312, 326. 332
331
Index
Shapiro, Harold N., 85, 146, 332 Shapiro’s Tauberian theorem, 85 Shen, MokKong, 9,332 Sierpinski, Waclaw, 11, 148, 332 Smallest primitive roots (table of), 213 Squarefree, 21 von Staudt, Karl Georg Christian, 275 Subgroup, 130 Summation formula of Euler, 54 Symbol, Jacobi (nip), 188 Legendre (n 1p). 179 System of residues, complete, 110 reduced. 113 Tatuzawa, Tikao, 99, 332 Tauberian theorem, 85 Theta function, 306 Titchmarsh, Edward Charles, 301, 332 Totient function cp(n), 25 Triangular numbers, 2, 326 Triangularnumber theorem, 326 Trivial zeros of i(s), 259 Twin primes, 6 Unique factorization theorem, 17 Uspensky, J. V.. 332
338
ValleePoussin, C. Vinogradov, A. I., Vinogradov, I. M., Visibility of lattice Voronoi, G., 59
J. de la, 9, 74, 332 11, 332 10,304,332 points, 62
Wahisz, Arnold, 91, 332 Waring. Edward. 306 Warinzs problem, 306 Williams, H. C.. 6, 332 Wilson, John, 116 Wilson’s theorem, I16 Wrathall, Claude P., 7, 332 Wright, E. M., 330 Wolstenholme’s theorem, I16 Yin, Wenlin,
IO, 332
Zarnke, C. R., 6,332 Zerofree regions of c(s), 29 1, 292 Zeros, of Lfunction L(s, x), 274 of Riemann’s zeta function c(s), 259, 274, 293 Zeta function, Hurwitz, 249 periodic, 257 Riemann, 9.249 Zuckerman. Herbert S.. 331